hdu 5113 Black And White （dfs回溯+剪枝）

时间：2015-01-23 23:05:18 阅读：229 评论：0 收藏：0 [点我收藏+]

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Black And White

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 854 Accepted Submission(s): 218
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

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

剪枝方法：若剩余空格数为 res ，对于每种颜色的数量t ，都要满足（res+1）> t 。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=15;

int ans[maxn][maxn],a[maxn],n,m,k,flag,t,xx,yy;

void input()
{
    flag=0;
    memset(ans,0,sizeof(ans));
    scanf("%d %d %d",&n,&m,&k);
    for(int i=1; i<=k; i++)   scanf("%d",&a[i]);
}

void dfs(int x,int y,int res)
{
    if(res==0)  flag=1;
    if(flag)  return ;
    for(int i=1;i<=k;i++)   if((res+1)/2<a[i])  return ;
    for(int i=1; i<=k; i++)
    {
        if(a[i] && ans[x-1][y]!=i && ans[x][y-1]!=i)
        {
            ans[x][y]=i,a[i]--;
            if(y+1>m)  xx=x+1,yy=1;
            else   xx=x,yy=y+1;
            dfs(xx,yy,res-1);
            if(flag)   return ;
            ans[x][y]=0,a[i]++;
        }
    }
}

void solve(int co)
{
    printf("Case #%d:\n",co);
    dfs(1,1,n*m);
    if(!flag)  printf("NO\n");
    else
    {
         printf("YES\n");
         for(int i=1;i<=n;i++)
         {
             for(int j=1;j<=m;j++)
             {
                 if(j==1)  printf("%d",ans[i][j]);
                 else  printf(" %d",ans[i][j]);
             }
             printf("\n");
         }
    }

}

int main()
{
    int T;
    scanf("%d",&T);
    for(int co=1; co<=T; co++)
    {
        input();
        solve(co);
    }
    return 0;
}

hdu 5113 Black And White （dfs回溯+剪枝）

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原文地址：http://blog.csdn.net/u012596172/article/details/43061061

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