Sicily 1823. Hardest Job Ever!

1823. Hardest Job Ever!

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

This is really hard, find the number of zeros in the number 12345678987654321!, here ‘!’ means factorial.
Well, I’m kidding. No one on earth may ever solve this successfully.
In fact, you are a spy, and you have stolen some top secret of the enemy, now you are to find a way which takes you least time to escape.
There are some many crosspoints and some many roads, even between two crosspoints there can be multiple roads. You are confused, but happily you have your laptop with you. 

Input

The first line contains number T (T<=10), the number of test cases.
Each test case begins with two integers n and m (1<=n<=200, 0<=m<=10000), number of crosspoints and number of roads respectively. Next m lines each has three integers i,j,k (i!=j, 1<=k<=10000), indicating there is a road of length k connecting city i and city j.
You can assume that the crosspoints are numbered from 1 to n. 1 is your starting point and you need to go to crosspoint n.
Roads are bidirectional. 

Output

For each test case, output one number, the shortest distance. If there is no path exist, output -1. 

Sample Input

12 11 2 3

Sample Output

3

// Problem#: 1013// Submission#: 3160682// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <algorithm>#include <iostream>#include <string>#include <stdio.h>#include <queue>#include <string.h>#include <vector>#include <iomanip>#include <map>#include <stack>#include <functional>#include <list>#include <cmath>using namespace std;#define MAX_POINT 205#define INF 99999999struct edge {    int to, cost;    edge(int t, int c) {        to = t;        cost = c;    }};int n;vector<edge> E[MAX_POINT];int d[MAX_POINT];void DIJKSTRA(int s) {    priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;    for (int i = 1; i <= n; d[i++] = INF);    d[s] = 0;    q.push(make_pair(0, s));    while (!q.empty()) {        pair<int, int> now = q.top();        q.pop();        if (d[now.second] < now.first) continue;        for (int i = 0; i < E[now.second].size(); i++) {            edge e = E[now.second][i];            if (d[e.to] > d[now.second] + e.cost) {                d[e.to] = d[now.second] + e.cost;                q.push(make_pair(d[e.to], e.to));            }        }    }}int main() {    std::ios::sync_with_stdio(false);        int caseNum;    cin >> caseNum;    while (caseNum--) {        int m;        cin >> n >> m;        for (int i = 0; i < m; i++) {            int from, to, cost;            cin >> from >> to >> cost;            E[from].push_back(edge(to, cost));            E[to].push_back(edge(from, cost));        }        DIJKSTRA(1);        if (d[n] == INF) cout << -1 << endl;        else cout << d[n] << endl;        for (int i = 0; i < n; i++) E[i].clear();    }    //getchar();    //getchar();        return 0;}