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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.

InputThe 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 .
OutputFor 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
DFS

 1 #include <iostream>
 2 using namespace std;
 3 #include<string.h>
 4 #include<set>
 5 #include<stdio.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<map>
 9 #include<algorithm>
10 #include<queue>
11 int a[30][30],yanse[30];
12 int lenx,leny,yanseshu;
13 int flag=0;
14 int b[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
15 int panduan(int x,int y,int k)
16 {
17     if(a[x-1][y]==k) return 0;
18     if(a[x][y-1]==k) return 0;
19     return 1;
20 }
21 int dfs(int x,int y)
22 {
23     if(x>leny)
24     return 1;
25     int shengyu=(leny-x)*lenx+lenx-y+2;
26     for(int i=1;i<=yanseshu;i++)
27     if(shengyu/2<yanse[i])
28     return 0;
29     for(int i=1;i<=yanseshu;i++)
30     {
31         int f=0;
32         if(yanse[i]&&panduan(x,y,i)){
33             a[x][y]=i;
34             yanse[i]--;
35             if(y==lenx)
36             f=dfs(x+1,1);
37             else
38              f=dfs(x,y+1);
39             yanse[i]++;
40         }
41         if(f)
42             return 1;
43     }
44     return 0;
45 }
46 int main()
47 {
48     int add=0,t;
49     cin>>t;
50     while(t--)
51     {
52         memset(a,0,sizeof(a));
53         memset(yanse,0,sizeof(yanse));
54         cin>>leny>>lenx>>yanseshu;
55         for(int i=1;i<=yanseshu;i++)
56             cin>>yanse[i];
57             cout<<"Case #"<<++add<<":"<<endl;
58             flag=0;
59             if(dfs(1,1))
60             {
61                 cout<<"YES"<<endl;
62                 for(int i=1;i<=leny;i++)
63                 {
64                     for(int j=1;j<=lenx;j++)
65                     {
66                         if(j==lenx)
67                         {
68                             cout<<a[i][j];
69                             continue;
70                         }
71                         cout<<a[i][j]<<‘ ‘;
72                     }
73                     cout<<endl;
74                 }
75             }
76             else
77                     cout<<"NO"<<endl;
78     }
79     return 0;
80 }

View Code

Black And White HDU - 5113