Description
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.
Output
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.
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
Source
SPFA的典型应用，不过就是需要两次建图，一次从站1出发，一次从其他站回来，这个时候逆向建图，然后两次调用SPFA算法，求得的最短路加起来就得到解了。
注意不要溢出。
不使用vector，stack这些STL容器果然快点的。
#include <stdio.h>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
#include <limits.h>
#include <string.h>
#include <stack>
#include <algorithm>
using namespace std;
const int MAX_N = 1000001;
struct Edge
{
int src, des, wei, next;
};
Edge eds[MAX_N];//edges' pool
int edsNum;//the total numbers of edges
int head[MAX_N];//the head edge indicator of all vertices
const int EDGE_END= -1;
void addEdge(int src, int des, int wei)
{
eds[edsNum].src = src;
eds[edsNum].des = des;
eds[edsNum].wei = wei;
eds[edsNum].next = head[src];
head[src] = edsNum++;
}
inline int nextEdge(int e)
{
return eds[e].next;
}
template<typename T>
inline bool equ(T a, T b) { return a == b; }
int dist[MAX_N], stk[MAX_N];
bool inStk[MAX_N];
void shortestSpanningTreeSPFA(int n)
{
fill(dist, dist+n+1, INT_MAX);
fill(inStk, inStk+n+1, false);
dist[1] = 0;
int top = -1;
stk[++top] = 1;
inStk[1] = true;
while (top != -1)
{
int u = stk[top--];
inStk[u] = false;
for (int e = head[u]; e != EDGE_END; e = nextEdge(e))
{
int v = eds[e].des, w = eds[e].wei;
if (dist[u] + w < dist[v])
{
dist[v] = dist[u] + w;
if (!inStk[v])
{
stk[++top] = v;
}
}
}
}
}
void reverseGrap(int V, int E)
{
fill(head, head+V+1, EDGE_END);
for (int e = 0; e < E; e++)
{
swap(eds[e].des, eds[e].src);
eds[e].next = head[eds[e].src];
head[eds[e].src] = e;
}
}
int main()
{
int T, V, E, u, v, w;
scanf("%d", &T);
while (T--)
{
scanf("%d %d", &V, &E);
fill(head, head+V+1, EDGE_END);
edsNum = 0;
for (int i = 0; i < E; i++)
{
scanf("%d %d %d", &u, &v, &w);
addEdge(u, v, w);//顶点下标由1开始
}
shortestSpanningTreeSPFA(V);
long long ans = 0LL;
for (int i = 1; i <= V; i++)
{
ans += dist[i];
}
reverseGrap(V, E);
shortestSpanningTreeSPFA(V);
for (int i = 1; i <= V; i++)
{
ans += dist[i];
}
printf("%lld\n", ans);
}
return 0;
}