Tour
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1148 Accepted Submission(s): 605
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.
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.
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 9999999
#define maxn 205
int lx[maxn],ly[maxn];
int w[maxn][maxn];
int Match[maxn],visx[maxn],visy[maxn];
int slack[maxn];
int ans,n,m;
bool findPath(int x)
{
int tmp;
visx[x]=1;
for(int y=1; y<=n; y++)
{
if(visy[y])continue;
if(w[x][y]==lx[x]+ly[y])
{
visy[y]=1;
if(!Match[y]||findPath(Match[y]))
{
Match[y]=x;
return true;
}
}
else
slack[y]=min(slack[y],w[x][y]-lx[x]-ly[y]);
}
return false;
}
void km()
{
memset(Match,0,sizeof(Match));
memset(ly,0,sizeof(ly));
for(int i=1; i<=n; i++)
{
lx[i]=INF;
for(int j=1; j<=n; j++)
lx[i]=min(lx[i],w[i][j]);
}
for(int x=1; x<=n; x++)
{
for(int i=1; i<=n; i++)
slack[i]=INF;
while(true)
{
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
if(findPath(x))break;
int tmp=INF;
for(int i=1; i<=n; i++)
{
if(!visy[i])
{
if(tmp>slack[i])
tmp=slack[i];
}
}
if(tmp==INF)return ;
for(int i=1; i<=n; i++)
{
if(visx[i])lx[i]+=tmp;
if(visy[i])ly[i]-=tmp;
}
}
}
}
int main()
{
int cas;
cin>>cas;
while(cas--)
{
int a,b,c;
cin>>n>>m;
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
w[i][j]=INF;
}
for(int i=0; i<m; i++)
{
cin>>a>>b>>c;
if(w[a][b]>c)
w[a][b]=c;
}
km();
int ans=0;
for(int i=1; i<=n; i++)
if(Match[i])
ans+=w[Match[i]][i];
cout<<ans<<endl;
}
return 0;
}