Description

As every one known, a football team has 11 players . Now, there is a big problem in front of the Coach Liu. The final contest is getting closer. 
Who is the center defense, the full back or the forward? ...... There are n wonderful players for n positions in the team and Coach Liu know 
everyone's abilities at different positions in matches. Assume that the team's power is the sum of the abilities of all n players according to 
their positions, could you help Coach Liu to find out the max power his team can get?

输入格式

The first line is an integer n(n<11). Followed by n rows. Each row has n integer (0 to 1000) which represents the abilities of one player 
at different positions.

输出格式

The max power.

输入样例

10
4 6 3 3 4 5 7 4 9 0
5 9 3 4 6 1 7 3 9 3
1 5 8 0 5 4 2 7 9 3
4 6 9 4 7 3 7 9 5 4
2 0 1 3 2 5 8 4 6 2
1 5 8 4 2 6 8 0 4 2
1 4 2 6 8 9 4 2 6 8
1 2 9 5 6 4 2 7 5 7
2 4 7 5 8 5 3 2 6 4
2 4 6 4 8 7 3 5 7 3

输出样例

76

简单回溯：wa在由于now是全局变量，在每一次到达结束状态时，now的值不会因递归返回而改变（不像局部变量），而我是希望在递归枚举所有情况是now的值应该是返回上一状态的，
故因  加一句：now-=mat[i][j]

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int mat[12][12];
int vis[12];
int now=0;
int maxr=0;
void dfs(int cur)
{
    int i;
    if(cur==n) {if(now>maxr) maxr=now;}
    else for(i=0;i<10;++i){
        if(!vis[i]){
            now+=mat[cur][i];//尝试选择第cur行第i列的数
            vis[i]=1;
            dfs(cur+1);
            vis[i]=0;
            now-=mat[cur][i];//谨记，now需随递归返回原来的值
        }

    }
}
int main()
{
    memset(vis,0,sizeof(vis));
    memset(mat,0,sizeof(mat));
    scanf("%d",&n);
    for(int i=0;i<n;++i)
        for(int j=0;j<n;++j)
        scanf("%d",&mat[i][j]);
        dfs(0);
        printf("%d
",maxr);
}