山东省第七届ACM竞赛 C题 Proxy （Dijkstra算法，单源路径最短问题）

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstring>


using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1000 + 100;
int n;
struct edge{
    int from, to, dist;
    edge(int u, int v, int d) : from(u), to(v), dist(d) { }
};
struct Headnode{
    int d, u;
    Headnode(int dd, int uu) : d(dd), u(uu) { }
    bool operator < (const Headnode &rhs) const {
        return d > rhs.d;
    }
};
struct Dijkstra{
    int m;
    vector<edge> edges;
    vector<int> G[maxn];
    bool done[maxn];
    int d[maxn];
    int p[maxn];

    void init(){
        for(int i = 0; i <= n+1; ++i)  G[i].clear();
        edges.clear();
    }

    void addedge(int f, int t, int d){
        edges.push_back(edge(f, t, d));
        m = edges.size();
        G[f].push_back(m-1);
    }

    void dijkstra(int s){
        priority_queue<Headnode> q;
        for(int i = 0; i <= n+1; ++i)  d[i] = INF;
        d[s] = 0;

        memset(done, 0, sizeof(done));
        q.push(Headnode(0, s));
        while(!q.empty()){
            Headnode x = q.top();  q.pop();
            int u = x.u;
            if(done[u])  continue;
            done[u] = true;
            for(int i = 0; i < G[u].size(); ++i){
                edge &e = edges[G[u][i]];
                if(d[e.to] >= d[u] + e.dist){
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = u;
                    q.push(Headnode(d[e.to], e.to));
                }
                else if(d[e.to] == d[u] + e.dist)  p[e.to] = min(G[u][i], p[e.to]);
            }
        }
    }
};
Dijkstra dijk;

int main(){
    int T, m;  cin >> T;
    while(T--){
        dijk.init();
        scanf("%d %d", &n, &m);
        int u, v, c;
        for(int i = 0; i < m; ++i){
            scanf("%d %d %d", &u, &v, &c);
            dijk.addedge(v, u, c);
        }
        dijk.dijkstra(n+1);

        if(dijk.d[0] >= INF)  printf("-1\n");
        else if(dijk.p[0] == n+1)  printf("0\n");
        else  printf("%d\n", dijk.p[0]);
    }
    return 0;
}