文章标题 HDU 5113 : Black And White （dfs+剪枝）

最新推荐文章于 2021-02-03 15:17:06 发布

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本文探讨了一个类似于四色定理的问题，即如何使用K种颜色给N×M的棋盘染色，使得相邻格子颜色不同且每种颜色使用次数固定。通过深度优先搜索（DFS）算法解决该问题，并给出实例解析。

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Black And White

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）

题意：给你一个n*m的一个表格，然后有 k 种颜色（k <= n*m），第 i 种颜色需要染表格数有c【i】个所以c1+c2+…+ck=n*m;
问有没有可能把表格涂上颜色且每个表格与旁边的表格颜色不一样。
分析: 因为n,m<=5,所以直接dfs就行了。其中有剪枝，单有一种颜色剩余数目比剩余的（格数+1）的一半还多，说明必有两个相邻的格子同一种颜色。直接返回。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int mp[10][10];
int n,m,k;
int color[30];
int flag=0;
bool judge (int x,int y,int p){//判断颜色如果在（x,y）上涂p颜色是否合法
    if (x-1>=0&&mp[x-1][y]==p) return false;
    if (y-1>=0&&mp[x][y-1]==p) return false;
    return true;
}
void dfs(int x,int y,int num){//num表示剩余的格子数
    if (num==0) {//如果格子数为零，说明条件成立
        flag=1;
        return;
    }
    for (int i=1;i<=k;i++){//剪枝
        if (color[i]>(num+1)/2) return;
    } 
    for (int i=1;i<=k;i++){
        if (color[i]>0&&judge(x,y,i)){
            color[i]--;
            mp[x][y]=i;
            if ((y+1)%m){//判断这一行是否已满，满的话进入下一行
                dfs(x,y+1,num-1);
            }
            else dfs(x+1,0,num-1);
            if (flag) return;//如果条件成立，返回
            color[i]++;//将减少的数目加回去
            mp[x][y]=0;
        }

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