hdoj 5113 Black And White DFS+剪枝

Black And White

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 929 Accepted Submission(s): 238
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 ci 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 ci (ci > 0), denoting the number of cells where the i-th color should be used.

It’s guaranteed that c1 + c2 + · · · + cK = 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

给你n*m的格子和K个颜色，然后问你能不能涂出每个格子相邻的颜色都不一样的图案，任意输出一种就好

题解

DFS，但是会T，所以需要神剪枝
如果当前的颜色数量，大于剩下格子的一半的话，那么肯定是可以直接return的，因为必然会有两个相同颜色的格子挨在一起
所以，我们可以就这样剪枝

代码

int c[maxn];
int flag=0;
int kiss[maxn][maxn];
int n,m,k;
void dfs(int x,int y,int cur)
{

	if(flag==1)
		return;
	if(cur==0)
	{
		flag=1;
		printf("YES
");
		REP_1(i,n)
		{
			REP_1(j,m)
			{
				if(j==1)
					printf("%d",kiss[i][j]);
				else
					printf(" %d",kiss[i][j]);
			}
			printf("
");
		}
		return;
	}

	REP_1(i,k)
	{
		if(c[i]>(cur+1)/2)
			return;
	}

	//cout<<x<<" "<<y<<" "<<cur<<endl;
	REP_1(i,k)
	{
		if((kiss[x-1][y]!=i&&kiss[x][y-1]!=i)&&c[i]>0)
		{
			kiss[x][y]=i;
			c[i]--;
			if(y==m)
				dfs(x+1,1,cur-1);
			else
				dfs(x,y+1,cur-1);
			c[i]++;
			kiss[x][y]=0;
		}
	}
}
int main()
{
	int t;
	RD(t);
	REP_1(ti,t)
	{
		RD(n),RD(m),RD(k);
		memset(c,0,sizeof(c));
		memset(kiss,0,sizeof(kiss));
		flag=0;
		REP_1(i,k)
			cin>>c[i];
		printf("Case #%d:
",ti);
		dfs(1,1,n*m);
		if(flag!=1)
			printf("NO
");
	}
}