USACO八皇后

VIS 0 1 2分别竖线和两个对角线,参见对角线(x,y)的和 差关系表

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=15;
int n,vis[5][N<<1],c[N],ans=0;
void dfs(int cur){
    if(cur==n+1){
        ans++;
        if(ans>3) return;
        for(int i=1;i<=n;i++) printf("%d ",c[i]); printf("\n");
        return;    
    }
    for(int i=1;i<=n;i++){
        if(!vis[0][i]&&!vis[1][i+cur]&&!vis[2][cur-i+n]){
            vis[0][i]=vis[1][i+cur]=vis[2][cur-i+n]=1;
            c[cur]=i;
            dfs(cur+1);
            vis[0][i]=vis[1][i+cur]=vis[2][cur-i+n]=0;
        }
    }
}
int main(){
    cin>>n;
    dfs(1);printf("%d",ans);
    return 0;
}

 

posted @ 2016-08-19 22:32  Candy?  阅读(235)  评论(0编辑  收藏  举报