(5113)HDU

本文根据http://www.cnblogs.com/fraud，这篇博客学习借鉴来的·····，知识产权很重要，我写博客的目的只是提升自己···做个笔记用···

Black And White

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 c ₁ + c ₂ + · · · + 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亚洲区北京站-重现赛（感谢北师和上交）

这个是杭电的题面，这道题就是问你是否能输出这个图形，一共有1~k个颜色，输出一个N*M的矩阵，这个矩阵的规则是每个颜色会有Ai个，在用完的前提下，保证相邻的颜色不相同。。。。差不多就是这个样子

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<set>
#include<algorithm>
#include<cmath>


#define ll __int64
#define MAX 1000009
using namespace std;


int k;
int ans[109][109];
int a[109];
int n,m;
int flag;


void dfs(int x,int y,int xx)
{
    if(!xx)//如果没有空位就会退回
    {
        flag = 1;
        return ;
    }
    for(int i = 1;i<=k;i++)//如果剩下的个数大于空位数，退出（剪支）
    {
        if(a[i]>(xx+1)/2)
            return ;
    }
    for(int i = 1;i<=k;i++)
    {
        if(!a[i])continue;//没有颜色跳过当前颜色循环
        if(x&&ans[x-1][y]==i)continue;//如果周围有当前颜色就会跳过
        if(y&&ans[x][y-1]==i)continue;//如果周围有当前颜色就会跳过
        a[i]--;//用掉一个
        ans[x][y] = i;//当前位置赋值成第i种颜色
        if(y<m-1)
            dfs(x,y+1,xx-1);//如果Y轴没有填满 先填Y轴
        else
            dfs(x+1,0,xx-1);//否则填X轴
        if(flag) return ;
        a[i]++;//不符合条件加回来
    }
    return ;
}


int main()
{
    int cas = 1;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        //memset(ans,0,sizeof(ans));
        flag = 0;
        scanf("%d%d%d",&n,&m,&k);
        for(int i = 1;i<=k;i++)scanf("%d",&a[i]);
        printf("Case #%d:\n",cas++);
        dfs(0,0,n*m);
        if(flag)
        {
            printf("YES\n");
            for(int i = 0;i<n;i++)
            {
                for(int j = 0;j<m;j++)
                {
                    if(j)
                        printf(" ");
                    printf("%d",ans[i][j]);
                }
                printf("\n");
            }
        }
        else
            printf("NO\n");
    }
    return 0;
}