Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 527    Accepted Submission(s): 145
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

·学习傅总思路打出来的代码，基本相同。

　　http://fudq.blog.163.com/blog/static/191350238201411271332692/

·一直还是比较怕dfs的，我只能说，，，练吧。。。。

·这题的剪枝也是醉了，现场想到了，加的位置不对还是没过，，，不知道是否是出题者故意的，，如果是就太神了。。

AC Code：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <string.h>
 6 #include <math.h>
 7 #include <queue>
 8 #include <stack>
 9 #include <stdlib.h>
10 #include <map>
11 using namespace std;
12 #define LL long long
13 #define sf(a) scanf("%d",&(a));
14
15 #define N 21
16 int r[N][N];
17 int n,m,flag,k;
18 int f[10010];
19 //染色：  dfs，深搜 + 剪枝
20
21
22 void Output(){
23     printf("YES\n");
24     for(int i=1;i<=n;i++){
25         //if(i==0) printf("%d",r[i][0]);
26         for(int j=1;j<=m;j++)
27             if(j==1)
28                 printf("%d",r[i][j]+1);
29             else printf(" %d",r[i][j]+1);
30         printf("\n");
31     }
32 }
33 int jud(int x,int y,int i){
34     if(r[x-1][y] == i) return 0;
35     if(r[x][y-1] == i) return 0; //表示此种方式不行！
36     return 1;
37 }
38 void dfs(int x,int y,int cnt){
39     if(flag) return ;
40     if(cnt == 0){
41         flag=1;
42         Output();
43         return ;
44     }
45     //加上剪枝
46     for(int i=0;i<k;i++){
47         if(f[i] > (cnt+1)/2) return ; //直接返回，此时不成功！
48     }
49     for(int i=0;i<k;i++){
50         if(f[i] && jud(x,y,i)){
51             r[x][y] = i;
52             f[i]--;
53             if(y==m) dfs(x+1,1,cnt-1);
54             else dfs(x,y+1,cnt-1);
55             r[x][y]=-1;
56             f[i]++;
57         }
58     }
59
60 }
61 int main(){
62     int T,cas=1;
63     scanf("%d",&T);
64     while(T--){
65         printf("Case #%d:\n",cas++);
66         scanf("%d %d %d",&n,&m,&k);
67         flag=0;
68         for(int i=0;i<k;i++) scanf("%d",&f[i]);
69         memset(r,-1,sizeof(r));
70         dfs(1,1,n*m);
71         if(!flag) printf("NO\n");
72     }
73
74     return 0;
75 }