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搜索（剪枝优化）：HDU 5113 Black And White

2024-08-16 02:45:22   221人阅读

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

41 5 24 13 3 41 2 2 42 3 32 2 23 2 32 2 2

Sample Output

Case #1:NOCase #2:YES4 3 42 1 24 3 4Case #3:YES1 2 32 3 1Case #4:YES1 22 33 1
　　这道题就是搜索，剪枝优化是如果当前未染色的点为x,(x+1)/2<max(col[i]),就可以return了。
 1 #include <algorithm> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdio> 5 using namespace std; 6 const int N=6; 7 int id[N][N],L[N*N],U[N*N],c[N*N]; 8 int T,cas,n,m,k,map[N*N],p[N*N]; 9 bool DFS(int x){10     if(x==n*m+1)return true;11     for(int i=1;i<=k;i++)12         if((n*m+2-x)/2<c[i])return false;13     for(int i=1;i<=k;i++){14         if(c[i]==0)continue;15         if(map[L[x]]!=i&&map[U[x]]!=i){16             c[i]-=1;map[x]=i;17             if(DFS(x+1))return true;18             c[i]+=1;map[x]=0;19         }20     }21     return false;22 }23 int main(){24     scanf("%d",&T);25     while(T--){int idx=0;26         scanf("%d%d%d",&n,&m,&k);27         for(int i=1;i<=k;i++)28             scanf("%d",&c[i]);    29         for(int i=1;i<=n;i++)30             for(int j=1;j<=m;j++)31                 id[i][j]=++idx;32         for(int i=1;i<=n;i++)33             for(int j=1;j<=m;j++){34                 L[id[i][j]]=id[i][j-1];35                 U[id[i][j]]=id[i-1][j];36             }37         printf("Case #%d:\n",++cas);38         if(DFS(1)){39             puts("YES");40             for(int i=1;i<=n;i++){41                 for(int j=1;j<m;j++)42                     printf("%d ",map[(i-1)*m+j]);43                 printf("%d\n",map[i*m]);    44             }    45         }    46         else puts("NO");            47     }48     return 0;49 }

 

搜索（剪枝优化）：HDU 5113 Black And White


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