HDU 5113 Black And White DFS+剪枝

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

题意：

一个n*m的数组，然后k种颜色，每种颜色有若干个（你可以把颜色存到一个数组里color【k】），然后相邻的方格颜色不同（共用一条边的方块颜色不同），

输出其中一种方案

只用DFS放颜色 是会超时的，所以我们需要一个剪枝，如果剩下的方格（就是还没有被染色的方格）的数目+1然后除以2，小于任意一种颜色的数目，那就没有必要继续往下搜了，（rem+1）/2<color[i],比如说3*3 的方格，如果你还剩的颜色中一种颜色是6个，那这种方案就不成立（体会一下）。因为9个方格中，颜色一样的最多有5个，

上代码，在代码中体会精髓

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=15;
int ans[maxn][maxn],color[maxn];
int n,m,k,flag,xx,yy;
void init(){//初始化
    flag=0;
	memset(ans,0,sizeof(ans));
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=k;i++)
	scanf("%d",&color[i]);
}
void dfs(int x,int y,int rem){
	if(rem==0) flag=1;
	if(flag==1) return ;
	for(int i=1;i<=k;i++)//很重要的一步剪枝
	if((rem+1)/2<color[i])//如果剩下的空白格子+1/2 小于任何一种颜色的数量，这种方案就不成立
	return ;
	for(int i=1;i<=k;i++)
	{
		if(color[i]&&ans[x-1][y]!=i&&ans[x][y-1]!=i)
		{
			ans[x][y]=i,color[i]--;
			if(y+1>m) xx=x+1,yy=1;
			else xx=x,yy=y+1;
			dfs(xx,yy,rem-1);
			if(flag)return ;
			ans[0][0]=0,color[i]++;
		}
	}
}
void solve(int e){
	printf("Case #%d:\n",e);
	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");
		}
	}
} 
int main(){
	int t;
	scanf("%d",&t);
	for(int e=1;e<=t;e++)
	{
		init();
		solve(e);
	}
	return 0;	
};