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划分树--查询区间k-th number code

一、pku_2104
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#include<stdio.h>
#include<algorithm>

#define N 100000+100
#define M 21  // log(N)

using namespace std;

struct Node {
	int val,id;
} a[N];

int n,m,tr[M][N];

int cmp(Node aa, Node bb) {
	if(aa.val < bb.val || (aa.val == bb.val && aa.id < bb.id)) return 1;
	return 0;
}

void build_tree(int dep, int s, int t)
{
	if(s >= t) return;
	
	int i,j,k,mid = (s+t)/2;
	j = s; k = mid+1;
	
	for(i=s;i<=t;i++)
	{
	    if(tr[dep][i] <= mid)
	        tr[dep+1][j++] = tr[dep][i];
	    else
	        tr[dep+1][k++] = tr[dep][i];

	    tr[dep][i] = j-1;
	}
	
	build_tree(dep+1, s, mid);
	build_tree(dep+1, mid+1, t);
}

int find_tree(int dep, int s, int t, int i, int j, int k)
{
	if(s == t) return s;
	int ci, mid = (s+t)/2;
	
	int v = tr[dep][j]-(s-1);
	if(i > s) v = tr[dep][j] - tr[dep][i-1];
	
	if(v >= k)
	{
	    ci = s; if(i > s) ci = tr[dep][i-1]+1;
	    return find_tree(dep+1, s, mid, ci, tr[dep][j], k);
	}
	else
	{
	    ci = mid+1; if(i > s) ci = mid+1 + (i-1)-tr[dep][i-1];
	    return find_tree(dep+1, mid+1, t, ci, mid+j-tr[dep][j], k-v);
	}
	return 0;
}

int main()
{
	int i,j,k,ans;
	while(scanf("%d %d",&n,&m) != EOF)
	{
	    for(i=1;i<=n;i++) {
	        scanf("%d",&a[i].val); a[i].id = i;
	    }
	    
	    sort(a+1, a+1+n, cmp);
	    for(i=1;i<=n;i++) tr[0][a[i].id] = i;

	    build_tree(0, 1, n);
	    
	    while(m--)
	    {
	        scanf("%d %d %d",&i,&j,&k);
	        ans = find_tree(0, 1, n, i, j, k);
	        printf("%d\n",a[ans].val);
	    }
	}
	return 0;
}
二、hdu_2665
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#include<stdio.h>
#include<algorithm>

#define N 100000+100
#define M 21  // log(N)

using namespace std;

struct Node {
	int val,id;
} a[N];

int n,m,tr[M][N];

int cmp(Node aa, Node bb) {
	if(aa.val < bb.val || (aa.val == bb.val && aa.id < bb.id)) return 1;
	return 0;
}

void build_tree(int dep, int s, int t)
{
	if(s >= t) return;
   
	int i,j,k,mid = (s+t)/2;
	j = s; k = mid+1;
   
	for(i=s;i<=t;i++)
	{
	    if(tr[dep][i] <= mid)
	        tr[dep+1][j++] = tr[dep][i];
	    else
	        tr[dep+1][k++] = tr[dep][i];

	    tr[dep][i] = j-1;
	}
   
	build_tree(dep+1, s, mid);
	build_tree(dep+1, mid+1, t);
}

int find_tree(int dep, int s, int t, int i, int j, int k)
{
	if(s == t) return s;
	int ci, mid = (s+t)/2;
   
	int v = tr[dep][j]-(s-1);
	if(i > s) v = tr[dep][j] - tr[dep][i-1];
   
	if(v >= k)
	{
	    ci = s; if(i > s) ci = tr[dep][i-1]+1;
	    return find_tree(dep+1, s, mid, ci, tr[dep][j], k);
	}
	else
	{
	    ci = mid+1; if(i > s) ci = mid+1 + (i-1)-tr[dep][i-1];
	    return find_tree(dep+1, mid+1, t, ci, mid+j-tr[dep][j], k-v);
	}
	return 0;
}

int main()
{
	int i,j,k,ans,T;
	scanf("%d",&T);
	while(T--)
	{
	    scanf("%d %d",&n,&m);
	    for(i=1;i<=n;i++) {
	        scanf("%d",&a[i].val); a[i].id = i;
	    }
	   
	    sort(a+1, a+1+n, cmp);
	    for(i=1;i<=n;i++) tr[0][a[i].id] = i;

	    build_tree(0, 1, n);
	   
	    while(m--)
	    {
	        scanf("%d %d %d",&i,&j,&k);
	        ans = find_tree(0, 1, n, i, j, k);
	        printf("%d\n",a[ans].val);
	    }
	}
	return 0;
}
三、hdu_3727
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#include<stdio.h>
#include<algorithm>

#define N 100000+100
#define M 21  // log(N)

typedef long long LL;

using namespace std;

struct Input {
	int sta,s,t,k;
} q[N+N+N];

struct Node {
	int val,id;
} a[N];

int n,m,tr[M][N],C[N],b[N];

int cmp(Node aa, Node bb) {
	if(aa.val < bb.val || (aa.val == bb.val && aa.id < bb.id)) return 1;
	return 0;
}

int lowbit(int x) {
	return x&(-x);
}

void change(int x, int y) {
	while(x <= n) {
	    C[x] += y; x += lowbit(x);
	}
}

int cal(int x) {
	int t=0;
	while(x > 0) {
	    t += C[x]; x -= lowbit(x);
	}
	return t;
}

void build_tree(int dep, int s, int t)
{
	if(s >= t) return;
   
	int i,j,k,mid = (s+t)/2;
	j = s; k = mid+1;
   
	for(i=s;i<=t;i++)
	{
	    if(tr[dep][i] <= mid)
	        tr[dep+1][j++] = tr[dep][i];
	    else
	        tr[dep+1][k++] = tr[dep][i];

	    tr[dep][i] = j-1;
	}
   
	build_tree(dep+1, s, mid);
	build_tree(dep+1, mid+1, t);
}

int find_tree(int dep, int s, int t, int i, int j, int k)
{
	if(s == t) return s;
	int ci, mid = (s+t)/2;
   
	int v = tr[dep][j]-(s-1);
	if(i > s) v = tr[dep][j] - tr[dep][i-1];
   
	if(v >= k)
	{
	    ci = s; if(i > s) ci = tr[dep][i-1]+1;
	    return find_tree(dep+1, s, mid, ci, tr[dep][j], k);
	}
	else
	{
	    ci = mid+1; if(i > s) ci = mid+1 + (i-1)-tr[dep][i-1];
	    return find_tree(dep+1, mid+1, t, ci, mid+j-tr[dep][j], k-v);
	}
	return 0;
}

int main()
{
	int i,k,T,low,up,mid,cas=0;
	LL ans[5];
	char ch[33];
   
	while(scanf("%d",&T) != EOF)
	{
	   
	    n = 0; m = 0;
	    for(i=1;i<=T;i++)
	    {
	        scanf("%s", ch);
	        if(ch[0] == 'I')
	        {
	            q[i].sta = 0;
	            n++; scanf("%d", &a[n].val); a[n].id = n;
	        }
	        else
	        {
	            q[i].sta = ch[6] - 48;
	            if(ch[6] == '1') scanf("%d %d %d",&q[i].s,&q[i].t,&q[i].k);
	            else
	                scanf("%d",&q[i].k);
	        }
	    }
	   
	    sort(a+1, a+1+n, cmp);
	    for(i=1;i<=n;i++)
	    {
	        b[a[i].id] = tr[0][a[i].id] = i;
	        C[i] = 0;
	    }
	    C[0] = 0;

	    build_tree(0, 1, n);
	   
	    ans[1] = ans[2] = ans[3] = 0;
	   
	    m = 0;
	   
	    for(i=1;i<=T;i++)
	    {
	        if(q[i].sta == 0)
	        {
	            m++;
	            change(b[m], 1);
	        }
	        else
	        if(q[i].sta == 1)
	        {
	            k = find_tree(0, 1, n, q[i].s, q[i].t, q[i].k);
	            ans[1] += a[k].val;
	        }
	        else
	        if(q[i].sta == 2)
	        {
	            low = 1; up = n;
	            while(low < up) {
	                mid = (low + up)/2;
	                if(a[mid].val < q[i].k) low = mid+1; else up = mid;
	            }
	            mid = (low + up)/2;
	            ans[2] += cal(mid);
	        }
	        else
	        if(q[i].sta == 3)
	        {
	            low = 1; up = n;
	            while(low < up) {
	                mid = (low + up)/2;
	                if(cal(mid) < q[i].k) low = mid+1; else up = mid;
	            }
	            mid = (low + up)/2;
	            ans[3] += a[mid].val;
	        }
	    }
	    cas++;
	    printf("Case %d:\n%lld\n%lld\n%lld\n",cas,ans[1],ans[2],ans[3]);
	}
	return 0;
}
四、hdu_3473
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#include<stdio.h>
#include<algorithm>

#define N 100000+100
#define M 21  // log(N)

typedef long long LL;

using namespace std;

struct Node {
	int val,id;
} a[N];

int n,m,pos,tr[M][N];
LL less,more, sum[M][N];

int cmp(Node aa, Node bb) {
	if(aa.val < bb.val || (aa.val == bb.val && aa.id < bb.id)) return 1;
	return 0;
}

void build_tree(int dep, int s, int t)
{
	if(s >= t) return;
   
	int i,j,k, mid = (s+t)/2;
	LL s1,s2;
   
	j = s; k = mid+1;
	s1 = s2 = 0;
   
	for(i=s;i<=t;i++)
	{
	    if(tr[dep][i] <= mid)
	    {
	        s1 += a[tr[dep][i]].val;
	        sum[dep][j] = s1;
	        tr[dep+1][j++] = tr[dep][i];
	    }
	    else
	    {
	        s2 += a[tr[dep][i]].val;
	        sum[dep][k] = s2;
	        tr[dep+1][k++] = tr[dep][i];
	    }

	    tr[dep][i] = j-1;
	}
   
	build_tree(dep+1, s, mid);
	build_tree(dep+1, mid+1, t);
}

void find_tree(int dep, int s, int t, int i, int j, int k)
{
	if(s == t) { pos = s; return ; }
	int ci,cj,  mid = (s+t)/2;
	LL s1,s2;
   
	int v = tr[dep][j]-(s-1);
	if(i > s) v = tr[dep][j] - tr[dep][i-1];
   
	if(v >= k)
	{
	    ci = s; if(i > s) ci = tr[dep][i-1]+1;
	    find_tree(dep+1, s, mid, ci, tr[dep][j], k);
	   
	   
	    if(i == s) ci = 0; else ci = (i-1)-tr[dep][i-1];
	    if(ci == 0) s1 = 0; else s1 = sum[dep][mid+ci];
	   
	    cj = j-tr[dep][j];
	    if(cj == 0) s2 = 0; else s2 = sum[dep][mid+cj];
	   
	    more += (s2-s1);
	}
	else
	{
	    ci = mid+1; if(i > s) ci = mid+1 + (i-1)-tr[dep][i-1];
	    find_tree(dep+1, mid+1, t, ci, mid+j-tr[dep][j], k-v);
	   
	   
	    if(i > s) ci = tr[dep][i-1]; else ci = s-1;
	    if(ci < s) s1 = 0; else s1 = sum[dep][ci];
	   
	    cj = tr[dep][j]; 
	    if(cj < s) s2 = 0; else s2 = sum[dep][cj];
	   
	    less += (s2-s1);
	}
}

int main()
{
	int i,j,k,T,cas=0;
	scanf("%d",&T);
	while(T--)
	{
	    scanf("%d",&n);
	    for(i=1;i<=n;i++) {
	        scanf("%d",&a[i].val); a[i].id = i;
	    }
	   
	    sort(a+1, a+1+n, cmp);
	   
	    sum[0][0] = 0;
	    for(i=1;i<=n;i++)
	    {
	        tr[0][a[i].id] = i;
	        sum[0][i] = sum[0][i-1] + a[i].val;
	    }

	    build_tree(0, 1, n);
	   
	    cas++;
	    printf("Case #%d:\n", cas);
	   
	    scanf("%d",&m);
	    while(m--)
	    {
	        scanf("%d %d",&i,&j);
	        i++; j++; k = (j-i+2)/2;
	        less = 0; more = 0;
	        find_tree(0, 1, n, i, j, k);
	       
	        //printf("%d %lld %lld\n",a[pos].val,less,more);
	       
	        printf("%lld\n", (LL)(k-1)*(LL)a[pos].val-less + more-(LL)(j-i+1-k)*(LL)a[pos].val);
	    }
	    printf("\n");
	}
	return 0;
}

划分树--查询区间k-th number

划分树 – 查询区间 k-th number

http://poj.org/problem?id=2104
http://acm.hdu.edu.cn/showproblem.php?pid=2665
http://acm.hdu.edu.cn/showproblem.php?pid=3727
http://acm.hdu.edu.cn/showproblem.php?pid=3473

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tr[ log(N) ][ N ]
1. 先对原来的数 稳定排序 ，tr[0][i] = 原先的数 a[i] 在排序后的位置。
2. dep = 0; s = 1; t = n;
3. 递归建树 
build_tree(dep, s, t)
{
    if(s >= t) return;
    mid = (s+t)/2;  j = s;  k = mid+1;
    for(i=s;i<=t;i++) if( tr[dep][i] <= mid )  tr[dep+1][j++] = tr[dep][i]; else tr[dep+1][k++] = tr[dep][i];
    //  把s 到 t 一分为二， s 到 t 的每个数 如果排序后它排在该区间的前半部分就移到下一层的前半部。
    //  如果要计算小于和大于 k-th number 的数的和要多算 sum[dep][x] 即 dep+1 层中 s 到 x(x<=mid) 的和 或 mid+1 到 x(x>mid) 的和。
    tr[dep][i] = j-1;  // s 到 t 区间， tr[dep][i] 记录 s 到 i 分到前半部分的最后位置
    build_tree(dep+1, s, mid);  build_tree(dep+1, mid+1, t);
}
4. 查找区间 [i,j] 中的 k-th number ，其中 1<=k<=j-i+1;
find_tree(dep, s, t, i, j, k)
{
    if(s == t) return s;
    v = i 到 j 中分到左边的数
    if(v >= k) return find_tree(dep+1, s, mid, ci, cj, k); // ci, cj 对应 i, j 分到前半部分的位置。 分到右半部分的和加到大于k-th number 上
    else    return find_tree(dep+1, mid+1, t, ci, cj, k-v); // 分到左半部分的和加到小于k-th number 上
 }
 
时间复杂度 O( n*log(n) 预处理， log(n) 查询 ) ，空间大小 n*log(n)
 
序列 ： 2 5 9 8 4 3 1
排序后  1 2 3 4 5 8 9
所以 原序列对应的最终位置为 2 5 7 6 4 3 1
 
        按最终位置分                                   指向分到前半部分的最后位置
tr[0][] = 2 5 7 6 4 3 1                        处理后 tr[0][] = 1 1 1 1 2 3 4
tr[1][] = 2 4 3 1 || 5 7 6                            tr[1][] = 1 1 1 2 || 5 5 6
tr[2][] = 2 1 || 4 3 || 5 6 || 7                      tr[2][] = 0 1 || 2 3 || 5 5 || 7
tr[3][] = 1 || 2 || 3 || 4 || 5 || 6

插头 DP code7-8

七、fzu_1977
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#include<stdio.h>

#define N 50000   // N = all
#define M 1600000  // M = 3^m

int n,m,can[33][33],last[33][33];

int all,val[33],r[N],e[M],p[N][13],b[33],w[33],cas=0;

long long dp[2][N];

int H[N][15];

void change()
{
	int i,j,tmp[33][33];
	for(i=1;i<=n;i++) for(j=1;j<=m;j++) tmp[m-j+1][i] = can[i][j];
	j = n; n = m; m = j;
	for(i=1;i<=n;i++)
	    for(j=1;j<=m;j++) can[i][j] = tmp[i][j];
}

void input()
{
	int i,j;
	char ch[33];

	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
	{
	    scanf("%s",ch);
	    for(j=1;j<=m;j++)
	        if(ch[j-1] == 'O') can[i][j] = 1;
	        else
	        if(ch[j-1] == '*') can[i][j] = 2;
	        else
	            can[i][j] = 0;
	}

	if(n < m) change();

	int k=0;
	for(i=1;i<=n;i++) for(j=1;j<=m;j++)
	{
	    if(can[i][j] == 1) k++;
	    last[i][j] = k;
	}
}

int ok(int kk)
{
	int i,l,a[15],c[15];
	for(i=1;i<=m+1;i++) { a[i] = kk%3; kk /= 3; b[i] = -1; }
	l = 0;
	for(i=1;i<=m+1;i++)
	    if(a[i] == 1) c[++l] = i;
	    else
	    if(a[i] == 2)
	    {
	        if(l == 0) return 0;
	        b[c[l]] = i; b[i] = c[l];
	        l--;
	    }
	if(l > 0) return 0;
	return 1;
}

void init()
{
	int i,j;
	val[1] = 1;
	for(i=2;i<=m+2;i++) val[i] = val[i-1]*3;

	all=0;
	for(i=0;i<val[m+2];i++)
	{
	    e[i] = -1;
	    if(ok(i) == 1)
	    {
	        e[i] = all; r[all] = i;
	        for(j=1;j<=m+1;j++) p[all][j] = b[j];
	        all++;
	    }
	}

	for(i=0;i<all;i++)
	    for(j=1;j<=m+1;j++)
	    {
	        H[i][j] = r[i]/val[j]%3;
	    }
}

void solve()
{
	int i,j,k,LL,UU,sta,u,y;

	long long ans = 0;

	u = 0;
	for(i=0;i<all;i++) dp[u][i] = 0; dp[u][0] = 1;

	for(i=1;i<=n;i++)
	    for(j=1;j<=m;j++)
	    {
	        y = u; u = 1-u;
	        if(j == 1)
	        {
	            for(k=all-1;k>=0;k--)
	            {
	                if(e[r[k]/3] >= 0) dp[y][k] = dp[y][e[r[k]/3]];
	                if(r[k]%3 != 0) dp[y][k] = 0;
	            }
	        }

	        for(k=0;k<all;k++) dp[u][k] = 0;

	        for(k=0;k<all;k++)
	        {
	            LL = H[k][j]; UU = H[k][j+1];

	            if(can[i][j] == 0)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    dp[u][k] = dp[y][k];
	                }
	                continue;
	            }

	            if(can[i][j] == 2)
	            {
	                if(LL == 0 && UU == 0)
	                    dp[u][k] += dp[y][k];
	            }

	            if(LL == 0 && UU == 0)
	            {
	                sta = r[k] + val[j] + val[j+1]*2;
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(LL == 0)
	            {
	                dp[u][k] += dp[y][k];

	                sta = r[k] - UU*val[j+1] + UU*val[j];
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(UU == 0)
	            {
	                dp[u][k] += dp[y][k];

	                sta = r[k] - LL*val[j] + LL*val[j+1];
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(LL == 1 && UU == 1)
	            {
	                if(p[k][j+1] > 0)
	                {
	                    sta = r[k]-val[j]-val[j+1]-val[p[k][j+1]];
	                    dp[u][e[sta]] += dp[y][k];
	                }
	            }
	            else
	            if(LL == 2 && UU == 2)
	            {
	                if(p[k][j] > 0)
	                {
	                    sta = r[k]-val[j]*2-val[j+1]*2+val[p[k][j]];
	                    dp[u][e[sta]] += dp[y][k];
	                }
	            }
	            else
	            if(LL == 2 && UU == 1)
	            {
	                sta = r[k] - 2*val[j]-val[j+1];
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(LL == 1 && UU == 2)
	            {
	                if(r[k]-val[j]-val[j+1]*2 == 0 && last[i][j] == last[n][m])
	                {
	                    ans += dp[y][k];
	                }
	            }
	        }
	    }
	cas++;
	printf("Case %d: %lld\n",cas,ans);
}

int main()
{
	int i,T;
	scanf("%d",&T);

	m = 12;
	init(); r[all] = 1000000000;

	while(T-- > 0)
	{
	    input();
	    for(i=0;r[i]<val[m+2];i++); all = i;
	    solve();
	}
	return 0;
}
八、pku_3133
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#include<stdio.h>

#define N 60000+100 // 3^(m+1)

int n,m,can[33][33];

int dp[2][N],H[N][13],val[33];

void solve()
{
	int i,j,k,LL,UU,all,sta,u,y;

	all = val[m+2]; u = 0;
	for(i=0;i<all;i++) dp[u][i] = 1000000;
	dp[u][0] = 0;

	for(i=1;i<=n;i++)
	    for(j=1;j<=m;j++)
	    {
	        y = u; u = 1-u;
	        if(j == 1)
	        {
	            for(k=all-1;k>=0;k--)
	            {
	                dp[y][k] = dp[y][k/3];
	                if(k%3 != 0) dp[y][k] = 1000000;
	            }
	        }

	        for(k=0;k<all;k++) dp[u][k] = 1000000;

	        for(k=0;k<all;k++)
	        {
	            LL = H[k][j]; UU = H[k][j+1];

	            if(can[i][j] == 1)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    dp[u][k] = dp[y][k];
	                }
	                continue;
	            }

	            if(can[i][j] == 0)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    if(dp[u][k] > dp[y][k]) dp[u][k] = dp[y][k];

	                    sta = k+val[j]+val[j+1];
	                    if(dp[u][sta] > dp[y][k] + 2) dp[u][sta] = dp[y][k]+2;

	                    sta = k+(val[j]+val[j+1])*2;
	                    if(dp[u][sta] > dp[y][k] + 2) dp[u][sta] = dp[y][k]+2;
	                }
	                else
	                if(LL == 0)
	                {
	                    if(dp[u][k] > dp[y][k]+1) dp[u][k] = dp[y][k]+1;

	                    sta = k-val[j+1]*UU+val[j]*UU;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if(UU == 0)
	                {
	                    if(dp[u][k] > dp[y][k]+1) dp[u][k] = dp[y][k]+1;

	                    sta = k-val[j]*LL+val[j+1]*LL;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if(LL == 1 && UU == 1)
	                {
	                    sta = k-val[j]-val[j+1];
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	                else
	                if(LL == 2 && UU == 2)
	                {
	                    sta = k-(val[j]+val[j+1])*2;
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	            }
	            else
	            if(can[i][j] == 2)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    sta = k+val[j];
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;

	                    sta = k+val[j+1];
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if((LL == 1 && UU == 0) || (LL == 0 && UU == 1))
	                {
	                    sta = k-LL*val[j]-UU*val[j+1];
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	            }
	            else
	            if(can[i][j] == 3)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    sta = k+val[j]*2;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;

	                    sta = k+val[j+1]*2;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if((LL == 2 && UU == 0) || (LL == 0 && UU == 2))
	                {
	                    sta = k-LL*val[j]-UU*val[j+1];
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	            }
	        }
	    }
	if(dp[u][0] == 1000000) dp[u][0] = 0;
	printf("%d\n",dp[u][0]);
}

int main()
{
	int i,j;

	val[1] = 1;
	for(i=2;i<=9+2;i++) val[i] = val[i-1]*3;
	for(i=0;i<val[9+2];i++) for(j=1;j<=9+1;j++) H[i][j] = i/val[j]%3;

	while(scanf("%d %d",&n,&m) != EOF)
	{
	    if(n == 0 && m == 0)break;

	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++) scanf("%d",&can[i][j]);

	    solve();
	}
	return 0;
}