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;
}

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