Black And White HDU - 5113 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.

InputThe 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 .
OutputFor 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

这个题明白哪里要剪枝的话就比较好写。

剪枝的地方是每种颜色剩余的数量必须小于等于（剩下格子数+1）/2

这个画画图就明白了。可是我就不知道怎么想到这一点的，而且不知道这一点为什么这么有效，加上剪枝才78ms，不加超时

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int mp[10][10];
int c[30];
 int n,m,k;
int dfs(int x,int y)
{
    if(x>=n)return 1;
    int flag;
    for(int i=1;i<=k;i++)
    {
        int res=(n-x-1)*m+m-y;
        if(c[i]>(res+1)/2)return 0;
    }
    for(int i=1;i<=k;i++)
    {
        flag=0;
        if(c[i]==0)continue;

        if(mp[x-1][y]!=i&&mp[x][y-1]!=i)
        {
            c[i]--;
            mp[x][y]=i;
            if(y==m-1)
                flag=dfs(x+1,0);
            else
                flag=dfs(x,y+1);
            c[i]++;
        }
        if(flag)return 1;
    }
    return 0;
}
int main()
{
    int t;
    cin>>t;
    int z=1;
    while(t--)
    {

        cin>>n>>m>>k;
       // memset(c,0,sizeof(c));
        for(int i=1;i<=k;i++)
            scanf("%d",&c[i]);
            printf("Case #%d:\n",z++);
        if(dfs(0,0))
        {
            printf("YES\n");
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m-1;j++)
                    printf("%d ",mp[i][j]);
                printf("%d\n",mp[i][m-1]);
            }
        }
        else printf("NO\n");
    }
    return 0;
}

Black And White HDU - 5113 dfs+剪枝

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