<span>第七届山东省省赛C Proxy（最短路）</span>

题意：

给出n个点和一些单向边，问从0到n+1

如果不能到则输出-1

如果能一步到则输出0

否则输出第一个到达的节点

如果两条路距离相等，则输出较小的节点

思路：

赛场上从前向后扫然后又向前推的，，，特别别扭

回来之后想了下，可以建反向边，从n+1走到0记录前驱就好了

/* ***********************************************
Author        :devil
Created Time  :2016/6/10 14:20:54
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
using namespace std;
const int N=1010;
const int inf=0x3f3f3f3f;
struct node
{
    int v,d;
    node(int a=0,int b=0):v(a),d(b){}
    bool operator < (const node &a) const
    {
        return d>a.d;
    }
};
struct edge
{
    int v,cost;
    edge(int a=0,int b=0):v(a),cost(b){}
};
vector<edge> eg[N];
bool vis[N];
int n,m,dis[N],pre[N];
void dijkstra(int x)
{
    memset(vis,false,sizeof(vis));
    for(int i=0; i<=n+1; i++)
        dis[i]=inf;
    priority_queue<node>q;
    while(!q.empty()) q.pop();
    dis[x]=0;
    q.push(node(x,0));
    node tmp;
    while(!q.empty())
    {
        tmp=q.top();
        q.pop();
        int u=tmp.v;
        if(vis[u]) continue;
        vis[u]=1;
        for(int i=0; i<eg[u].size(); i++)
        {
            int v=eg[tmp.v][i].v;
            int cost=eg[u][i].cost;
            if(!vis[v]&&dis[v]>dis[u]+cost)
            {
                dis[v]=dis[u]+cost;
                pre[v]=u;
                q.push(node(v,dis[v]));
            }
            else if(!vis[v]&&dis[v]==dis[u]+cost)
                pre[v]=min(pre[v],u);
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,u,v,w;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n+1;i++)
            eg[i].clear();
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&w);
            eg[v].push_back(edge(u,w));
        }
        dijkstra(n+1);
        if(dis[0]==inf)
        {
            printf("-1\n");
            continue;
        }
        if(pre[0]==n+1)
        {
            printf("0\n");
            continue;
        }
        printf("%d\n",pre[0]);
    }
    return 0;
}