In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color.  — Wikipedia, the free encyclopedia  In this problem, you have to solve the 4-color problem. Hey, I’m just joking.  You are asked to solve a similar problem:  Color an N × M chessboard with K colors numbered from 1 to K such that no two adjacent cells have the same color (two cells are adjacent if they share an edge). The i-th color should be used in exactly c i cells.  Matt hopes you can tell him a possible coloring.

Input

The first line contains only one integer T (1 ≤ T ≤ 5000), which indicates the number of test cases.  For each test case, the first line contains three integers: N, M, K (0 < N, M ≤ 5, 0 < K ≤ N × M ).  The second line contains K integers c i (c i > 0), denoting the number of cells where the i-th color should be used.  It’s guaranteed that c 1 + c 2 + · · · + c K = N × M . 

Output

For each test case, the first line contains “Case #x:”, where x is the case number (starting from 1).  In the second line, output “NO” if there is no coloring satisfying the requirements. Otherwise, output “YES” in one line. Each of the following N lines contains M numbers seperated by single whitespace, denoting the color of the cells.  If there are multiple solutions, output any of them.

Sample Input

4
1 5 2
4 1
3 3 4
1 2 2 4
2 3 3
2 2 2
3 2 3
2 2 2

Sample Output

Case #1:
NO
Case #2:
YES
4 3 4
2 1 2
4 3 4
Case #3:
YES
1 2 3
2 3 1
Case #4:
YES
1 2
2 3
3 1

題意：給出一個n*m的棋盤，再給你k種顏色，每種顏色給你c個，這些顏色一共可以用n*m次，現在開始使用這些顏色將棋盤塗色相鄰的棋盤不能塗相同的顏色，並且要把顏色用完，不行就輸出NO，可以的話輸出YES和輸出這個圖

我看了下部落格是dfs+剪枝，後來自己寫了下過了，還好不是很難想。

剪枝主要就是每次看下剩下的格子的一半有沒有顏色中的一個多，因為是不能相鄰塗相同顏色，那麼剩下格子數的一半cnt與剩下的每種顏色對比，如果剩下一種顏色的個數大於cnt，那麼就一定塗不完。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int c[50],mapp[50][50];
int flag,n,m,k;
void dfs(int x,int y,int cnt)
{
    if(cnt==0)
    {
        flag=1;
        return ;
    }
    if(flag)
        return ;
    for(int i=1;i<=k;i++)//剪枝
    {
        if((cnt+1)/2<c[i])  return ;
    }
    int xx,yy;
    for(int i=1;i<=k;i++)
    {
        if(c[i]&&mapp[x-1][y]!=i&&mapp[x][y-1]!=i)
        {
            mapp[x][y]=i;
            c[i]--;
            if(y+1>m)
                yy=1,xx=x+1;
            else yy=y+1,xx=x;
            dfs(xx,yy,cnt-1);
            if(flag)
                return ;
            mapp[x][y]=0;
            c[i]++;
        }
    }
}
int main()
{
    int t,casee=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=k;i++)
            scanf("%d",&c[i]);
        memset(mapp,0,sizeof(mapp));
//        printf("%d\n",n*m);
        flag=0;
        dfs(1,1,n*m);
        printf("Case #%d:\n",++casee);
        if(flag==0)
            printf("NO\n");
        else
        {
            printf("YES\n");
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=m;j++)
                {
                    if(j!=1)
                        printf(" ");
                    printf("%d",mapp[i][j]);
                }
                printf("\n");
            }
        }

    }
}