Light OJ 1019

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1019

题目大意：Tanvir想从节点1的位置走到节点n的位置， 输出最短距离， 如果不存在输出"Impossible".

解题思路：dijkstra模版题

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
 
const int INF = 0x3f3f3f3f;
const int N = 103;
int dis[N];
vector<pair<int, int> > vec[N];
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > >que;
 
void dijkstra()
{
    dis[1] = 0;
    que.push(make_pair(0, 1));
    while(!que.empty())
    {
        pair<int, int> p = que.top();
        que.pop();
        int v = p.second;
        if(dis[v] < p.first)
            continue;
 
        for(int i=0; i<vec[v].size(); ++ i)
        {
            pair<int, int> t = vec[v][i];
            if(dis[t.second] > dis[v] + t.first)
            {
                dis[t.second] = dis[v] + t.first;
                que.push(make_pair(dis[t.second], t.second));
            }
        }
    }
}
void solve(int cases)
{
    for(int i=1; i<=100; ++ i)
    {
        dis[i] = INF;
        vec[i].clear();
    }
 
    while(!que.empty())
        que.pop();
 
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=1; i<=m; ++ i)
    {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        vec[u].push_back(make_pair(w, v));
        vec[v].push_back(make_pair(w, u));
    }
 
    dijkstra();
    if(dis[n] == 0x3f3f3f3f)
        printf("Case %d: Impossible
", cases);
    else
        printf("Case %d: %d
", cases, dis[n]);
}
 
int main()
{
    int T;
    scanf("%d", &T);
    for(int i=1; i<=T; ++ i)
        solve(i);
    return 0;
}