HDU 3488 Tour //2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

 

Tour

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 380    Accepted Submission(s): 196

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
   
   
   
   

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

 

 

Recommend

zhouzeyong

 

 

全图构成哈密顿回路，这种题目已经做过很多，不多说了 #include<cstdio> #include<cstring> #include<vector> using namespace std; const int inf=1<<28; int g[1005][1005];//邻接矩阵 int lx[1005],ly[1005];//顶点标号 bool sx[1005],sy[1005];//是否已经搜索过 int link[1005],n; vector<int> q[1010]; int min(int a,int b) { if(a<b) return a; return b; } bool path(int k) { sx[k]=true; for(int i=1; i<=n; i++) { if(!sy[i]&&(lx[k]+ly[i]==g[k][i])) { sy[i]=1; if(link[i]==-1||path(link[i])) { link[i]=k; return true; } } } return false; } int BestMatch() { int d,sum; memset(ly,0,sizeof(ly)); memset(link,-1,sizeof(link)); for(int i=1; i<=n; i++) { lx[i]=-inf; for(int j=1; j<=n; j++) if(lx[i]<g[i][j]) lx[i]=g[i][j]; } for(int k=1; k<=n; k++) { while(1) { memset(sx,0,sizeof(sx)); memset(sy,0,sizeof(sy)); if(path(k)) break;//匹配成功 d=inf; for(int i=1; i<=n; i++) if(sx[i]) for(int j=1; j<=n; j++) if(!sy[j]) d=min(d,lx[i]+ly[j]-g[i][j]); for(int i=1; i<=n; i++) { if(sx[i]) lx[i]-=d; if(sy[i]) ly[i]+=d; } } } sum=0; for(int i=1; i<=n; i++) { if(link[i]==-1||g[link[i]][i]==-inf) return -1; sum+=g[link[i]][i]; } return -sum; } int main() { int m,T,cas=1; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) q[i].clear(); for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) g[i][j]=inf; for(int i=1; i<=m; i++) { int x,y,c; scanf("%d%d%d",&x,&y,&c); g[x][y]=min(g[x][y],c); } for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) g[i][j]=-g[i][j]; int t=BestMatch(); printf("%d/n",t); } return 0; }