HDU 5113 Black And White

Black And White

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 5859 Accepted Submission(s): 1615
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）

Recommend

liuyiding

题解：

搜索+减枝优化（判断是否有颜色的数量超过剩下的方格数量的一半）

参考代码：

 1 #include <bits/stdc++.h>
 2 #define INF 0x3f3f3f3f
 3 #define eps 1e-9
 4 #define rep(i,a,n) for(int i=a;i<n;++i)
 5 #define per(i,a,n) for(int i=n-1;i>=a;--i)
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 #define np next_permutation
10 #define mp make_pair
11 using namespace std;
12 const int maxn=1e5+5;
13 const int maxm=1e5+5;
14 
15 int t,n,m,k,flag;
16 //dfs 找方案剪枝
17 int dp[10][10],c[100];
18 bool check1(int x,int y,int i)
19 {
20     if(x-1>=1 && dp[x-1][y]==i) return false;
21     if(y-1>=1 && dp[x][y-1]==i) return false;
22     return true;
23 }
24 void dfs(int x,int y,int le)
25 {
26     rep(i,1,k+1) if(c[i]>(le+1)/2) return;
27     if(flag) return;
28     if(x==n+1) { flag=1; return; }
29     for(int i=1;i<=k;i++)
30     {
31         if(c[i])
32         {
33             if(check1(x,y,i))
34             {
35                 c[i]--;
36                 dp[x][y] = i;
37                 if(y==m) dfs(x+1,1,le-1);
38                 else dfs(x,y+1,le-1);
39                 if(flag) return;
40                 c[i]++; dp[x][y] = 0;
41             }
42         }
43     }
44 } 
45 int main()
46 {
47     scanf("%d",&t);
48     rep(tt,1,t+1)
49     {
50         flag=0;
51         memset(dp,0,sizeof(dp));
52         scanf("%d%d%d",&n,&m,&k);
53         rep(i,1,k+1) scanf("%d",c+i);
54         dfs(1,1,n*m);
55         printf("Case #%d:
",tt);
56         if(flag)
57         {
58             printf("YES");
59             rep(i,1,n+1)
60             {
61                 rep(j,1,m) printf("%d ",dp[i][j]);
62                 printf("%d
",dp[i][m]);
63             }
64         }
65         else printf("YES");
66     }
67     return 0;
68 } 
69

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