HDOJ 5113 Black And White DFS+剪枝

DFS+剪枝...

在每次DFS前,當前棋盤的格子數量的一半小於一種顏色的數量時就剪掉

Black And White

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 194 Accepted Submission(s): 50
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

Source 2014ACM/ICPC亞洲區北京站-重現賽（感謝北師和上交）

#include 
#include 
#include 
#include 

using namespace std;

int n,m,k;
int mp[6][6];

struct CC
{
    int num,color;
}c[30];

int color[30];
int ccc[30];

bool cmp(CC a,CC b)
{
    return a.num=0&&x=0&&y(cnt+1)/2) return false;
	}
	for(int i=0;i=m) {ny=0; nx++;}
			dfs(nx,ny,cnt-1);
			if(flag==true) return true;
			ccc[color[i]]++;
			mp[x][y]=0;
			vis[i]=false;
		}
	}
	return false;
}

int main()
{
    int T_T,cas=1;
    scanf("%d",&T_T);
    while(T_T--)
    {
        memset(mp,0,sizeof(mp));
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i(((n+1)/2)*((m+1)/2)+(n/2)*(m/2)))
        {
            puts("NO"); continue;
        }
        int pos=0;
        for(int i=0;i