2014ACM/ICPC亚洲区北京站-B - Black And White

HDU - 5113

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

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

dfs+剪枝

剪枝方法：若剩余空格数为 res ，对于每种颜色的数量t ，都要满足（res+1）> t 。

#include<bits/stdc++.h>
using namespace std;
const int maxn=15;
 
int ans[maxn][maxn],a[maxn],n,m,k,flag,t,xx,yy;
void input()
{
    flag=0;
    memset(ans,0,sizeof(ans));
    scanf("%d %d %d",&n,&m,&k);
    for(int i=1; i<=k; i++)   scanf("%d",&a[i]);
}
 
void dfs(int x,int y,int res)
{
    if(res==0) flag=1;
    if(flag) return ;
    for(int i=1;i<=k;i++)  
	    if((res+1)/2<a[i]) return;
    for(int i=1;i<=k;i++)
    {
        if(a[i]&&ans[x-1][y]!=i&&ans[x][y-1]!=i)
        {
            ans[x][y]=i,a[i]--;
            if(y+1>m)  xx=x+1,yy=1;
            else   xx=x,yy=y+1;
            dfs(xx,yy,res-1);
            if(flag)   return ;
            ans[x][y]=0,a[i]++;
        }
    }
}
 
int main()
{
    int T;
    scanf("%d",&T);
    for(int co=1; co<=T; co++)
    {
        input();
		printf("Case #%d:\n",co);
        dfs(1,1,n*m);
        if(!flag)  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("%d",ans[i][j]);
                    else  printf(" %d",ans[i][j]);
                }
                printf("\n");
            }
        }
    }
    return 0;
}

2014ACM/ICPC亚洲区北京站-B - Black And White

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