Black And White(DFS+剪枝)

Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3937    Accepted Submission(s): 1082
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

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:第一行一个 T ，然后3个整数 n,m,k 代表有个 n 行 m 列的矩阵，k 代表有k种颜色，然后是 k 种颜色分别有多少个

要用这些颜色涂满矩阵，相邻颜色不能相同，问能否填出

 

DFS+剪枝，剪枝就是，如果有一种颜色比剩下的格子+1的一半还多的话，就 return

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <vector>
using namespace std;
#define LL long long
#define PI acos(-1.0)
#define lowbit(x) (x&(-x))
#define INF 0x7f7f7f7f      // 21 E
#define MEM 0x7f            // memset 都变为 INF
#define MOD 4999            // 模
#define eps 1e-9            // 精度
#define MX  10         // 数据范围

int read() {    //输入外挂
    int res = 0, flag = 0;
    char ch;
    if((ch = getchar()) == '-') flag = 1;
    else if(ch >= '0' && ch <= '9') res = ch - '0';
    while((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0');
    return flag ? -res : res;
}
// code... ...
int n,m,k;
int ok;
int color[MX*MX];
int num[MX][MX];

void dfs(int x,int y)
{
    if (x>n) ok=1;
    for (int i=1;i<=k;i++)
    {
        int remain = n*m-((x-1)*m+y-1)+1;
        if (color[i]>remain/2) return;
    }
    for (int i=1;i<=k;i++)
    {
        if (color[i]>0&&num[x-1][y]!=i&&num[x][y-1]!=i)
        {
            num[x][y]=i;
            color[i]--;
            if (y==m) dfs(x+1,1);
            else dfs(x,y+1);
            color[i]++;
            if (ok) return;
        }
    }
}

int main()
{
    int T=read();
    for (int cnt=1;cnt<=T;cnt++)
    {
        n=read();m=read();k=read();
        for (int i=1;i<=k;i++)
            color[i]=read();
        ok = 0;
        memset(num,0,sizeof(num));
        dfs(1,1);
        printf("Case #%d:\n",cnt);
        if (ok)
        {
            printf("YES\n");
            for (int i=1;i<=n;i++)
            {
                for (int j=1;j<m;j++)
                    printf("%d ",num[i][j]);
                printf("%d\n",num[i][m]);
            }
        }
        else printf("NO\n");
    }
    return 0;
}