Lightoj 1019 - Brush (V)

算出从点1到点n的最短路径。

/* ***********************************************
Author        :guanjun
Created Time  :2016/6/23 22:58:19
File Name     :1019.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int dis[][];
int n,m;
int lowcost[];
int vis[];
int dij(){
    cle(vis);
    vis[]=;
    memset(lowcost,INF,sizeof lowcost);
    for(int i=;i<=n;i++)lowcost[i]=dis[][i];
    lowcost[]=;
    int k,Min;
    for(int i=;i<=n;i++){
        Min=INF;
        for(int j=;j<=n;j++){
            if(lowcost[j]<Min&&!vis[j])Min=lowcost[j],k=j;
        }
        if(Min==INF)break;
        vis[k]=;
        for(int j=;j<=n;j++){
            if(lowcost[j]>lowcost[k]+dis[k][j]&&!vis[j])
                lowcost[j]=lowcost[k]+dis[k][j];
        }
    }
    return lowcost[n];
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int T,x,y;
    ll z;
    cin>>T;
    for(int t=;t<=T;t++){
        cin>>n>>m;
        memset(dis,INF,sizeof dis);
        for(int i=;i<=m;i++){
            cin>>x>>y>>z;
            if(dis[x][y]>z)dis[y][x]=dis[x][y]=z;
        }
        int ans=dij();
        if(ans==INF){
            printf("Case %d: Impossible\n",t);
        }
        else printf("Case %d: %d\n",t,ans);

    }
    return ;
}