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\n",maxr);

}

巴特西

17111 Football team

Description

输入格式

输出格式

输入样例

输出样例

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