POJ1511

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.

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:

The
input consists of N cases. The first line of the input contains only
positive integer N. Then follow the cases. Each case begins with a line
containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is
the number of stops including CCS and Q the number of bus lines. Then
there are Q lines, each describing one bus line. Each of the lines
contains exactly three numbers - the originating stop, the destination
stop and the price. The CCS is designated by number 1. Prices are
positive integers the sum of which is smaller than 1000000000. You can
also assume it is always possible to get from any stop to any other
stop.

Sample Input:

For each
case, print one line containing the minimum amount of money to be paid
each day by ACM for the travel costs of its volunteers.

题目大意： 给你一张有向网，从1出发到每个节点并返回，问最小权值和。

这个题思维上的难点是返回时的权值计算，其实我们可以转换一下思维，返回时建立一张反向图然后再算即可。

此题可用（dijkstra+heap）或者SPFA做，我是用dijkstra+heap做的

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define fi first
#define se second
using namespace std;
#define gamma 0.5772156649015328606065120 //欧拉常数
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define N 1000010
typedef long long LL;
typedef pair<int,int> PII;

struct Road{
int next;
int to;
int v;
}road[N<<1];                   //邻接表存图，因为要反向构图一次，所以*2
int d
,n,h[2]
,cnt,vis
;
long long ans;

struct cmp{
bool operator()(const int &x,const int &y)const{
return d[x]>d[y];
}
};                                //从小到大

void init(int &x,int &y,int &v,int k){
road[cnt].to=y;
road[cnt].v=v;
road[cnt].next=h[k][x];
h[k][x]=cnt++;
}

void dijkstra(int k){
int u,i,pos;
memset(d,inf,sizeof(d));
memset(vis,0,sizeof(vis));
d[1]=0;
priority_queue<int,vector<int>,cmp>q;
q.push(1);
while(!q.empty()){
u=q.top(); q.pop();
vis[u]=1;
for(i=h[k][u]; i!=-1; i=road[i].next){
pos=road[i].to;
if(!vis[pos]&&d[pos]>d[u]+road[i].v){
d[pos]=d[u]+road[i].v;
q.push(pos);
}
}
}
}

int main(){
int group,m,x,y,v;
//freopen("lxx.txt","r",stdin);
scanf("%d",&group);
while(group--){
cnt=0;
memset(h,-1,sizeof(h));
scanf("%d%d",&n,&m);
for(int i=0; i<m; ++i){
scanf("%d%d%d",&x,&y,&v);
init(x,y,v,0);
init(y,x,v,1);  //反向构图
}
ans=0;
dijkstra(0);
for(int i=2; i<=n; ++i)
ans+=d[i];
dijkstra(1);
for(int i=2; i<=n; ++i)
ans+=d[i];
printf("%lld\n",ans);
}
return 0;
}