Best match in Bipartite Graph-8616 Auto Rally match

8616 Auto Rally

Time Limit: 500 ms memory limit: 1000 K

Submissions: 3 passes: 3

Question type: Programming Language: Unlimited

Description

The scau team is going to participate in the rally competition. In a rally, each car requires a driver and a navigator. There are n drivers and N navigators in the scau team, each of which has a tacit understanding of each navigator. Now we need to find a certain pair of N pairs, this maximizes the sum of the tacit values of the N pairs.

Input

Multiple cases. The first line of each case is an integer n and the range is (1 <= n <= 16 ). Next there are n rows, each row has n numbers, and the J number in line I represents the tacit understanding of the I driver and the J navigator. There are multiple groups of test data. When n = 0, the data ends and this line does not need to be processed.Output

Output the maximum value of the sum of tacit values after pairing.

Sample Input

1 5 2 10 20 25 5 0  

Sample output

5 45  

Source

CPP

This is obviously the best match for the bipartite graph. Obviously, I won't. I decided to find the information. After understanding and researching the standard process, I have used it as my own. This is the legendary km algorithm, understanding is a bit painful, because I often talk about what the tree is, and I don't know what it is, but I don't know what it is. I can think of the Code directly. below is my code...

#include<string>#include<algorithm>#define N 100using namespace std;int gp[N][N],vx[N],vy[N],fx[N],fy[N],n,m[N];int dfs(int u){vx[u]=1;for(int i=0;i<n;i++){if(vy[i]==0&&fx[u]+fy[i]==gp[u][i]){vy[i]=1;if(m[i]==-1||dfs(m[i])){m[i]=u;return 1;}}}return 0;}int count(){int i,j,k,d,ans;for(i=0;i<n;i++){fx[i]=-0x7fffffff;fy[i]=0;for(j=0;j<n;j++){if(fx[i]<gp[i][j]){fx[i]=gp[i][j];}}}memset(m,-1,sizeof(m));for(int u=0;u<n;u++){memset(vx,0,sizeof(vx));memset(vy,0,sizeof(vy));while(!dfs(u)){d=0x7fffffff;for(i=0;i<n;i++){if(vx[i])for(j=0;j<n;j++){if(!vy[j]){if(d>(fx[i]+fy[j]-gp[i][j]))d=fx[i]+fy[j]-gp[i][j];}}}for(i=0;i<n;i++){if(vx[i])fx[i]-=d,vx[i]=0;if(vy[i])fy[i]+=d,vy[i]=0;}}}ans=0;for(i=0;i<n;i++){ans+=gp[m[i]][i];}return ans;}int main(){int i,j;while(1){scanf("%d",&n);if(n==0)return 0;for(i=0;i<n;i++)for(j=0;j<n;j++){scanf("%d",&gp[i][j]);}printf("%d\n",count());}return 0;}