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=6;
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+1)return true;
    for(int i=1;i<=k;i++)
        if((n*m+2-x)/2<c[i])return false;
    for(int i=1;i<=k;i++){
        if(c[i]==0)continue;
        if(map[L[x]]!=i&&map[U[x]]!=i){
            c[i]-=1;map[x]=i;
            if(DFS(x+1))return true;
            c[i]+=1;map[x]=0;
        }
    }
    return false;
}
int main(){
    scanf("%d",&T);
    while(T--){int idx=0;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=k;i++)
            scanf("%d",&c[i]);    
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                id[i][j]=++idx;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++){
                L[id[i][j]]=id[i][j-1];
                U[id[i][j]]=id[i-1][j];
            }
        printf("Case #%d:
",++cas);
        if(DFS(1)){
            puts("YES");
            for(int i=1;i<=n;i++){
                for(int j=1;j<m;j++)
                    printf("%d ",map[(i-1)*m+j]);
                printf("%d
",map[i*m]);    
            }    
        }    
        else puts("NO");            
    }
    return 0;
}