Black And White

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
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

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:

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

题意：有一个n*m个地图，用k中颜色来进行填充，每种颜色可以使用的次数为c_i次，∑c_i=n*m,要求相邻的格子的颜色不能相同，问是否存在满足要求的染色方案，若存在，则输出其中一种。

分析：注意到n,m≤5,图较小，考虑用dfs来搞，但是光是dfs会T，所以需要加上一个剪枝。

若当前某种颜色的剩余数目大于剩余格子数目的一半，则必定不能完成填充方案，直接跳出。

 1 //gaoshenbaoyou  ------ pass system test
 2 #include <iostream>
 3 #include <sstream>
 4 #include <ios>
 5 #include <iomanip>
 6 #include <functional>
 7 #include <algorithm>
 8 #include <vector>
 9 #include <string>
10 #include <list>
11 #include <queue>
12 #include <deque>
13 #include <stack>
14 #include <set>
15 #include <map>
16 #include <cstdio>
17 #include <cstdlib>
18 #include <cmath>
19 #include <cstring>
20 #include <climits>
21 #include <cctype>
22 using namespace std;
23 #define XINF INT_MAX
24 #define INF 0x3FFFFFFF
25 #define MP(X,Y) make_pair(X,Y)
26 #define PB(X) push_back(X)
27 #define REP(X,N) for(int X=0;X<N;X++)
28 #define REP2(X,L,R) for(int X=L;X<=R;X++)
29 #define DEP(X,R,L) for(int X=R;X>=L;X--)
30 #define CLR(A,X) memset(A,X,sizeof(A))
31 #define IT iterator
32 typedef long long ll;
33 typedef pair<int,int> PII;
34 typedef vector<PII> VII;
35 typedef vector<int> VI;
36 const int maxn=30;
37 int a[maxn];
38 bool flag=0;
39 int ans[10][10];
40 int n,m,k;
41 void dfs(int x,int y,int left)
42 {
43     if(!left)
44     {
45         flag=1;
46         return;
47     }
48     for(int i=1;i<=k;i++)
49         if(a[i]>(left+1)/2)return;
50     for(int i=1;i<=k;i++)
51     {
52         if(!a[i])continue;
53         if(x&&ans[x-1][y]==i)continue;
54         if(y&&ans[x][y-1]==i)continue;
55         a[i]--;
56         ans[x][y]=i;
57         if(y<m-1)dfs(x,y+1,left-1);
58         else dfs(x+1,0,left-1);
59         if(flag)return;
60         a[i]++;    
61     }
62     return;
63 }
64 int main()
65 {
66     //ios::sync_with_stdio(false);
67     int t;
68     scanf("%d",&t);
69     int cas=1;
70     while(t--)
71     {
72         flag=0;
73         scanf("%d%d%d",&n,&m,&k);
74         int sum=0;
75         int maxx=0;
76         int tot=n*m;
77         for(int i=1;i<=k;i++)
78             scanf("%d",&a[i]);
79         printf("Case #%d:
",cas++);
80         dfs(0,0,tot);
81         if(flag)
82         {
83             printf("YES
");
84             for(int i=0;i<n;i++)
85             {
86                 for(int j=0;j<m;j++)
87                 {
88                     if(j)printf(" ");
89                     printf("%d",ans[i][j]);
90                 }
91                 printf("
");
92             }
93         }
94         else 
95             printf("NO
");
96     }
97     return 0;
98 }

代码君