The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.
All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.
Input
Output
Sample Input
2
2 2
1 2 13
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50
Sample Output
46
210 题意：求原点到各个点再回到原点的路径和
思路：直接从原点跑最短路，然后反向建图，再从原点跑一次最短路
#include<cstdio>
#include<cstring>
#include<queue> using namespace std; typedef pair<int,int>p;
typedef long long ll;
struct Node
{
int y,next;
int val;
}node[][]; int cnt[],head[][]; void add(int x,int y,int val,int t)
{
node[t][++cnt[t]].y=y;
node[t][cnt[t]].val=val;
node[t][cnt[t]].next=head[t][x];
head[t][x]=cnt[t];
}
int t;
bool vis[];
ll dist[][];
priority_queue<p,vector<p>,greater<p> >que;
void dijstra(int t)
{
memset(vis,,sizeof(vis));
while(!que.empty())que.pop();
memset(dist[t],0x3f,sizeof(dist[t]));
que.push(p(,));
while(!que.empty())
{
p tmp = que.top();
que.pop();
int s=tmp.second;
int v=tmp.first;
if(vis[s])continue;
vis[s] = ;
dist[t][s]= v;
for(int i=head[t][s];i;i=node[t][i].next)
{
int to=node[t][i].y;
if(dist[t][to] > 1ll*(v+node[t][i].val))
{
que.push(p(v+node[t][i].val,to));
}
}
}
}
int main()
{
scanf("%d",&t);
while(t--)
{
int p,q;
scanf("%d%d",&p,&q);
cnt[]=cnt[]=;
memset(head,,sizeof(head));
for(int i=;i<=q;i++)
{
int u,v,val;
scanf("%d%d%d",&u,&v,&val);
add(u,v,val,);
add(v,u,val,);
}
dijstra();
dijstra();
ll ans = ;
for(int i=;i<=p;i++)
{
ans += dist[][i] + dist[][i];
}
printf("%lld\n",ans);
}
}