搜索（剪枝优化）：HDU 5113 Black And White

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 c ₁ + c ₂ + ・ ・ ・ + 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
　　这道题就是搜索，剪枝优化是如果当前未染色的点数为x,(x+1)/2<max(col[i]),就可以return了。

 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int N=;
 int id[N][N],L[N*N],U[N*N],c[N*N];
 int T,cas,n,m,k,map[N*N],p[N*N];
 bool DFS(int x){
     if(x==n*m+)return true;
     for(int i=;i<=k;i++)
         if((n*m+-x)/<c[i])return false;
     for(int i=;i<=k;i++){
         if(c[i]==)continue;
         if(map[L[x]]!=i&&map[U[x]]!=i){
             c[i]-=;map[x]=i;
             if(DFS(x+))return true;
             c[i]+=;map[x]=;
         }
     }
     return false;
 }
 int main(){
     scanf("%d",&T);
     while(T--){int idx=;
         scanf("%d%d%d",&n,&m,&k);
         for(int i=;i<=k;i++)
             scanf("%d",&c[i]);
         for(int i=;i<=n;i++)
             for(int j=;j<=m;j++)
                 id[i][j]=++idx;
         for(int i=;i<=n;i++)
             for(int j=;j<=m;j++){
                 L[id[i][j]]=id[i][j-];
                 U[id[i][j]]=id[i-][j];
             }
         printf("Case #%d:\n",++cas);
         if(DFS()){
             puts("YES");
             for(int i=;i<=n;i++){
                 for(int j=;j<m;j++)
                     printf("%d ",map[(i-)*m+j]);
                 printf("%d\n",map[i*m]);
             }
         }
         else puts("NO");
     }
     return ;
 }