HDU5113 Black And White

Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 4198    Accepted Submission(s): 1148
Special Judge

Problem Description

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 c₁ + c₂ + · · · + 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

 
题目大意是说给你一个N*M的矩阵，再给你K个整数填进这个矩阵，输入Ci(1<=i<=K),Ci表示第i个数字的个数。
要剪枝

#include<cstring>
#include<cstdio>
int n,m,k;
int array[26];
int map[30][30];
int xnext,ynext;
int count=1;
int flag;
void dfs(int x,int y,int z){
	if(z==n*m){
		flag=1;
		printf("Case #%d:\n",count++);
		printf("YES\n");
		for(int i=1;i<=n;i++){
			for(int j=1;j<=m;j++){
				if(j==m){
					printf("%d\n",map[i][j]);
					break;
				}
				printf("%d ",map[i][j]);
			}
		}
		return;
	}
	else{
		for(int i=1;i<=k;i++){
			if((n*m-z+1)/2<array[i])//剪枝
			return;
		}
		for(int i=1;i<=k;i++){
			if(array[i]!=0&&map[x-1][y]!=i&&map[x][y-1]!=i){
				array[i]--;
				map[x][y]=i;
				if(y+1>m){
					xnext=x+1;
					ynext=1;
				}
				else{
					xnext=x;
					ynext=y+1;
				}
				dfs(xnext,ynext,z+1);
				if(flag) return;
				map[x][y]=0;
				array[i]++;
			}
		}
	}
}
int main(){
	int t;
	scanf("%d",&t);
	int ans=0;
	while(t--){
		scanf("%d%d%d",&n,&m,&k);
		flag=0;
		for(int i=1;i<=k;i++)
			scanf("%d",&array[i]);
		memset(map,0,4);
		dfs(1,1,0);
		if(flag==0){
			printf("Case #%d:\n",count++);
			printf("NO\n");
		}	
	}
}