UVa 10917 A Walk through the forest

題目大意：
給你n個點m條邊的無向圖， 問從起點（1號點）到達終點（2號點）的總路徑數量（要求從每個點u走到下一個點v時滿足：v到終點的距離小於u到終點的距離。
分析：
從起點出發到每個點的最短路徑可以用一次Dijkstra算法求出，那麼原問題就轉化成了在一個Dag上求路徑數目，可以直接DP或者利用Dijkstra計算出的雖短路樹來求解
代碼：

#include<bits/stdc++.h>
using namespace std;
const int inf = 0x7fffffff;
const int maxn = 1000 + 10;

struct edge {
    int from, to, dist;
};

struct Heapnode {
    int d, u;
    bool operator < (const Heapnode& rhs) const {
        return d > rhs.d;
    }
};

struct Dijkstra {
    int n, m;
    vector<edge> edges;
    vector<int> G[maxn];
    bool done[maxn]; 
    int dis[maxn], pre[maxn];

    void init(int n) {
        this->n = n;
        for(int i=0; i<n; i++) G[i].clear();
        edges.clear();
    }
    void add_edge(int from, int to, int dist) {
        edges.push_back((edge){from, to, dist});
        m = edges.size();
        G[from].push_back(m-1);
    }
    void dijkstra(int s) {
        priority_queue<Heapnode> q;
        memset(pre, 0, sizeof(pre));
        memset(done, false, sizeof(done));
        for(int i=0; i<n; i++) dis[i] = inf; dis[s] = 0;

        q.push((Heapnode){0, s}); 
        while(!q.empty()) {
            Heapnode x = q.top(); q.pop();
            int u = x.u; if(done[u]) continue; done[u] = true;

            for(int i=0; i<(int)G[u].size(); i++) {
                edge& e = edges[G[u][i]];
                if(dis[e.to] > dis[u] + e.dist) {
                    dis[e.to] = dis[u] + e.dist;
                    pre[e.to] = G[u][i];
                    q.push((Heapnode){dis[e.to], e.to});
                }
            }
        }
    }
}Dij;

int n, m, u, v, w;
int ways[maxn];

int dfs(int u) {
    if(u == 1) return 1;
    int& ans = ways[u];
    if(ans >= 0) return ans;

    ans = 0;
    for(int i=0; i<(int)Dij.G[u].size(); i++) {
        int v = Dij.edges[Dij.G[u][i]].to;
        if(Dij.dis[u] > Dij.dis[v]) ans += dfs(v);
    }
    return ans;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    freopen("ans.txt", "w", stdout);
#endif
    while(scanf("%d%d", &n, &m) == 2 && n+m) {
        Dij.init(n);
        for(int i=0; i<m; i++) {
            scanf("%d%d%d", &u, &v, &w); u--, v--;
            Dij.add_edge(u, v, w);
            Dij.add_edge(v, u, w);
        }
        Dij.dijkstra(1);
        memset(ways, -1, sizeof(ways));
        printf("%d\n", dfs(0));
    }
    return 0;
}