uva167 - The Sultan's Successors

题意：八皇后问题的扩展。8*8棋盘上每个格子都有一个整数，要求8个皇后所在格子的数字之后最大

解法一，回溯：

用vis数组记录列，主对角（y-x），副对角（y+x）访问情况

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int C[50], vis[3][50], tot = 0, n = 8, nc = 0;
int board[8][8];
int max_score;

void search(int cur)
{
    nc++;
    if(cur==n)
    {
        tot++;
        int score=0;
        for(int i=0;i<8;i++)
            score+=board[i][C[i]];
        max_score=max(max_score, score);
    }
    else for(int i=0;i<n;i++)
    {
        if(vis[0][i] || vis[1][i-cur+n] || vis[2][i+cur])
            continue;
        C[cur]=i;
        vis[0][i]=vis[1][i-cur+n]=vis[2][i+cur]=1;
        search(cur+1);
        vis[0][i]=vis[1][i-cur+n]=vis[2][i+cur]=0;
    }
}

int main()
{
    int k;
    scanf("%d", &k);
    while(k--)
    {
        for(int i=0;i<8;i++)
            for(int j=0;j<8;j++)
            {
                scanf("%d", &board[i][j]);
            }
        memset(vis, 0, sizeof(vis));
        max_score=0;
        search(0);
        printf("%5d\n", max_score);
    }
    return 0;
}

解法二， STL next_permutation 遍历所有列排列

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int v[8][8], P[8];

bool judge() {
    for(int i=0;i<8;i++)
        for(int j=0;j<i;j++)
            if(j-P[j]==i-P[i] || j+P[j]==i+P[i])
                return false;
    return true;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--) {
        for(int i=0;i<8;i++) {
            P[i]=i;
            for(int j=0;j<8;j++)
                scanf("%d", &v[i][j]);
        }
        int ans=0;
        do
        {
            if(!judge())
                continue;
            int score=0;
            for(int i=0;i<8;i++) score+=v[i][P[i]];
            ans=max(ans, score);
        }while(next_permutation(P, P+8));
        printf("%5d\n", ans);
    }
    return 0;
}

uva167 - The Sultan's Successors

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