HDU 5113--Black And White(搜索+剪枝)

题目链接

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

题意：有一个N*M的方格板，现在要在上面的每个方格上涂颜色，有K种颜色，每种颜色分别涂c[1]次、c[2]次……c[K]次，c[1]+c[2]+……+c[K]=N*M

要求每个方格的颜色与其上下左右均不同，如果可以输出YES，并且输出其中的一种涂法，如果不行，输出NO；

思路：暴力搜索，但是这样会超时，可以在搜索中加入剪枝：对于剩余的方格数res，以及当前剩余的颜色可涂数必须满足（res+1）/2>=c[i]

否则在当前情况下继续向下搜得不到正确涂法；

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int N,K,M;
int c[];
int mp[][];

int check(int x,int y,int k)
{
    int f=;
    if(mp[x-][y]==k) f=;
    if(mp[x][y-]==k) f=;
    return f;
}

int cal(int x,int y)
{
    if(x>N) return ;
    int res=(N-x)*M+M-y+; ///剩余方格数+1 ；
    for(int i=;i<=K;i++) if(res/<c[i]) return ; ///剪枝，某种颜色剩余方格数>(剩余方格数+1)/2 肯定不对；
    for(int i=;i<=K;i++)
    {
        int f=;
        if(c[i]&&check(x,y,i)){
            mp[x][y]=i; c[i]--;
            if(y==M)  f=cal(x+,);
            else      f=cal(x,y+);
            c[i]++;
        }
        if(f) return ;
    }
    return ;
}

int main()
{
    int T,Case=;
    cin>>T;
    while(T--)
    {
       scanf("%d%d%d",&N,&M,&K);
       for(int i=;i<=K;i++) scanf("%d",&c[i]);
       printf("Case #%d:\n",Case++);

       if(!cal(,)) { puts("NO"); continue; }
       puts("YES");
       for(int i=;i<=N;i++)
       for(int j=;j<=M;j++)
         printf("%d%c",mp[i][j],(j==M)?'\n':' ');
    }
    return ;
}