HDOJ5113 Black And White【dfs+剪枝】

最新推荐文章于 2022-06-19 16:21:10 发布

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最新推荐文章于 2022-06-19 16:21:10 发布

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探讨了数学中的四色定理，并提出了一种类似的问题，即如何使用K种颜色对N×M的棋盘进行染色，使得相邻格子颜色不同，且每种颜色使用次数固定。

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Black And White

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1383 Accepted Submission(s): 366
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

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：给一个n*m的网格给K种颜色和每种颜色的数量求是否能将这些颜色放入网格中且相邻两个格子颜色不同；直接dfs超时；需要剪枝当某种颜色的数量大于剩余格子的一半时必定有两种颜色相邻

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int map[6][6];
int color[26];
int n,m,k,flag;
bool judge(int x,int y,int i)
{
	if(color[i]==0)return false;
	if(map[x][y-1]==i||map[x-1][y]==i)return false;
	return true;
}
void dfs(int x,int y)
{
	int i,j;
	if(flag)return;
	if(y>m){
		dfs(x+1,1);
		return;
	}
	if(x>n)
	{
		flag=1;printf("YES\n");
		for(i=1;i<=n;++i)
		{
			for(j=1;j<=m;++j)
			{
				if(j==1)printf("%d",map[i][j]);
				else
				printf(" %d",map[i][j]);
			}
			printf("\n");
		}
		return;
	}
	for(i=1;i<=k;++i)
	{
		if(color[i]>((n*m-(x-1)*m+y)/2))return;
	}
	for(i=1;i<=k;++i)
	{
		if(judge(x,y,i))
		{
			map[x][y]=i;
			color[i]--;
			dfs(x,y+1);
			color[i]++;
		}
	}
}
int main()
{
	int t,i,j,x=1,max;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&n,&m,&k);
		max=-1;
		for(i=1;i<=k;++i)
		{
			scanf("%d",&color[i]);
			max=max>color[i]?max:color[i];
		}
		printf("Case #%d:\n",x++);
		if(max>n*m)
		{
			printf("NO\n");
			continue;
		}
		memset(map,0,sizeof(map));
		flag=0;
		dfs(1,1);
		if(flag==0)printf("NO\n");
	}
	return 0;
}