K.Teamwork Brings Profits! 典型的dfs搜索

K.Teamwork Brings Profits!

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 117   Accepted Submission(s) : 53

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Problem Description

Williammed's father is running a small company now, which has N employees.(2<=N<=10 and N is an even number). The employees are numbered from 1 to N. All the employees are divided into N/2 groups. One employee can only be in one group, and each group works on one project. As we all know, teamwork is very important to a company, so different team can make different profits. And now, given any two people i and j, Williammed's father can tell how much profit(Pij) can they make if they work together. Here comes the problem, given all the Pij(1 <= i <= N,1 <= j <= N,0 < Pij<= 100), you should tell the most profits this company can make. This is an easy problem, isn't it?

Input

The first line of the input is N(2<=N<=10 and N is an even number),the number of employees in the company.
Then there're N lines,each line has N numbers.The jth number in the ith line is Pij,as we discribe above.And we guarantee Pij = Pji,Pii = 0.
The end-of-file is denoted by a single line containing the integer 0.

Output

For each case,output the most profits this company can make.

Sample Input

40 6 62 136 0 35 9462 35 0 513 94 5 00

Sample Output

156

Author

syu

Source

Developing School's Contest 6

post code:

直接进行dfs搜索就可以了

#include<stdio.h>#include<string.h>int a[12][12];int visit[12];int max,t,sum,n;void dfs( int time ,int sum){  if(time==t){if(max<sum)max=sum;return;}  int i,j;   for(i=1;i<=n-1;i++)    for(j=i+1;j<=n;j++)    {      if(visit[i]==false&&visit[j]==false)      {       visit[i]=visit[j]=true; //记录此点已经搜索过了       dfs(time+1,sum+a[i][j]);        visit[i]=visit[j]=false;  //将此点变成未搜索过。         }    }                  }int main(){    int i,j;    while(scanf("%d",&n)&&n!=0)    {      max=0;       for(i=1;i<=n;i++)         for(j=1;j<=n;j++)          scanf("%d",&a[i][j]);          t=n/2;       memset(visit,false,sizeof(visit));              dfs(0,0);       printf("%d\n",max);                          }             }