Transportation Costs 
Total Submit: 129 Accepted: 34
Description
Minya Konka decided to go to Fuzhou to participate in the ACM regional contest at their own expense.Through the efforts, they got a small amount of financial support from their school, but the school could pay only one of those costs between two stations for them.
From SWUST to Fujian Normal University which is the contest organizer, there are a lot of transfer station and there are many routes.For example, you can take the bus to Mianyang Railway Station (airport), then take the train (plane) to Fuzhou,and then take a bus or taxi to the Fujian Normal University.
The school could pay only one of those costs between two stations for them, the others paid by the Minya Konka team members.They want to know what is the minimum cost the need pay.Can you calculate the minimum cost?
Input
There are several test cases.
In each case,the first line has two integers n(n<=100) and m(m<=500),it means there are n stations and m undirected roads.
The next m lines, each line has 3 integers u,v,w,it means there is a undirected road between u and v and it cost w.(1<=u,v<=n,0<=w<=100)
The ID of SWUST is 1,and n is the ID of Fujian Normal University.
Output
If they can not reach the destination output -1, otherwise output the minimum cost.
Sample Input
5 5
1 2 7
1 3 3
3 4 3
2 5 3
4 5 3
Sample Output
3
Source
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <cstring>
using namespace std;
const int INF = 0x7fffffff;
const int maxn = 110;
int gn, gm;
vector<pair<int, int> > g[maxn+10];
bool inque[maxn+10];
queue<int> Q;
void spfa(int s, int d[]) {
int i;
for(i = 1; i < maxn; i++) d[i] = INF;
d[s] = 0;
while(!Q.empty()) Q.pop();
Q.push(s);
inque[s] = true;
while(!Q.empty()) {
int u = Q.front();
Q.pop();
for(i = 0; i < (int)g[u].size(); i++) {
int t = g[u][i].first;
if(d[u] + g[u][i].second < d[t]) {
d[t] = d[u] + g[u][i].second;
if(!inque[t]) {
inque[t] = true;
Q.push(t);
}
}
}
inque[u] = false;
}
}
void init() {
int i;
for(i = 1; i <= gn; i++) {
g[i].clear();
}
}
void work(int d1[], int d2[]) {//枚举每一条边.
int i, j;
int mindis = INF;
int x;
for(i = 1; i <= gn; i++) {
for(j = 0; j < (int)g[i].size(); j++) {
x = g[i][j].first;
if(d1[i] != INF && d2[i] != INF && d1[i] + d2[x] < mindis) {
mindis = d1[i] + d2[x];
}
}
}
if(mindis != INF) {
printf("%d
", mindis);
}
else
printf("-1
");
}
int main()
{
int i;
int u, v, w;
int d1[maxn];
int d2[maxn];
pair<int, int> t;
while(scanf("%d%d", &gn, &gm) != EOF) {
init(); //清空容器.
for(i = 1; i <= gm; i++) {
scanf("%d%d%d", &u, &v, &w);
t.first = v;
t.second = w;
g[u].push_back(t);
t.first = u;
t.second = w;
g[v].push_back(t);
}
spfa(1, d1);//求起点到每个顶点的最短路径.
spfa(gn, d2);//求终点到每个顶点的最短路径.
work(d1, d2);//枚举每条边，求最小值.
}
return 0;
}