图论(二分图,KM算法)：HDU 3488 Tour

Tour

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2628    Accepted Submission(s): 1285

Problem Description

In
the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M
(M <= 30000) one-way roads connecting them. You are lucky enough to
have a chance to have a tour in the kingdom. The route should be
designed as: The route should contain one or more loops. (A loop is a
route like: A->B->……->P->A.)
Every city should be just in one route.
A
loop should have at least two cities. In one route, each city should be
visited just once. (The only exception is that the first and the last
city should be the same and this city is visited twice.)
The total distance the N roads you have chosen should be minimized.

 

Input

An integer T in the first line indicates the number of the test cases.
In
each test case, the first line contains two integers N and M,
indicating the number of the cities and the one-way roads. Then M lines
followed, each line has three integers U, V and W (0 < W <=
10000), indicating that there is a road from U to V, with the distance
of W.
It is guaranteed that at least one valid arrangement of the tour is existed.
A blank line is followed after each test case.

 

Output

For each test case, output a line with exactly one integer, which is the minimum total distance.

 

Sample Input

1

6 9

1 2 5

2 3 5

3 1 10

3 4 12

4 1 8

4 6 11

5 4 7

5 6 9

6 5 4

 

Sample Output

42

 

　　 这题就是KM算法的模板。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 const int INF=;
 int w[maxn][maxn],sx[maxn],sy[maxn],lx[maxn],ly[maxn];
 int match[maxn],slack[maxn];
 int n,m;

 bool Search(int x){
     sx[x]=true;
     for(int y=;y<=n;y++){
         if(sy[y])continue;//?
         int t=lx[x]+ly[y]-w[x][y];
         if(t)
             slack[y]=min(slack[y],t);
         else{
             sy[y]=true;
             if(!match[y]||Search(match[y])){
                 match[y]=x;
                 return true;
             }
         }
     }
     return false;
 }

 int KM(){
     memset(lx,0x80,sizeof(lx));
     memset(ly,,sizeof(ly));
     for(int i=;i<=n;i++)
         for(int j=;j<=n;j++)
             lx[i]=max(lx[i],w[i][j]);

     for(int i=;i<=n;i++){
         memset(slack,,sizeof(slack));
         while(true){
             memset(sx,,sizeof(sx));
             memset(sy,,sizeof(sy));
             if(Search(i))
                 break;
             int minn=INF;
             for(int j=;j<=n;j++)
                 if(!sy[j]&&minn>slack[j])
                     minn=slack[j];

             for(int j=;j<=n;j++)
                 if(sx[j])
                     lx[j]-=minn;

             for(int j=;j<=n;j++)
                 if(sy[j])
                     ly[j]+=minn;
                 else
                     slack[j]-=minn;
         }
     }
     int ret=;
     for(int i=;i<=n;i++)
         ret+=w[match[i]][i];
     return ret;
 }
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&n,&m);
         memset(w,0x80,sizeof(w));
         memset(match,,sizeof(match));
         for(int i=,a,b,c;i<=m;i++){
             scanf("%d%d%d",&a,&b,&c);
             w[a][b]=max(w[a][b],-c);
         }
         printf("%d\n",-KM());
     }
     return ;
 }