Emergency

题意：有N个点，M条边，每个点有权值，问从起点到终点最短路的个数以及权值最大的最短路的权值。

分析：修改Dijstra模板。

#include<bits/stdc++.h>
using namespace std;
const int INT_INF = 0x3f3f3f3f;
const int MAXN = 500 + 10;
int weight[MAXN];
typedef long long LL;
struct Edge{
    int from, to;
    LL dist;
    Edge(int f, int t, LL d):from(f), to(t), dist(d){}
};
struct HeapNode{
    LL d;
    int u;
    HeapNode(LL dd, int uu):d(dd), u(uu){}
    bool operator < (const HeapNode& rhs)const{
        return d > rhs.d;
    }
};
struct Dijkstra{
    int n, m;
    vector<Edge> edges;
    vector<int> G[MAXN];
    LL d[MAXN];
    int num[MAXN];
    int value[MAXN];
    bool done[MAXN];
    void init(int n){
        this -> n = n;
        for(int i = 0; i <= n; ++i) G[i].clear();
        edges.clear();
    }
    void AddEdge(int from, int to, LL 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;
        for(int i = 0; i <= n; ++i){
            d[i] = 0x3f3f3f3f3f3f3f3f;
        }
        memset(done, false, sizeof done);
        d[s] = 0;
        num[s] = 1;
        value[s] = weight[s];
        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 < 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;
                    num[e.to] = num[u];
                    value[e.to] = value[u] + weight[e.to];
                    Q.push(HeapNode(d[e.to], e.to));
                }
                else if(d[e.to] == d[u] + e.dist){
                    num[e.to] += num[u];
                    if(value[u] + weight[e.to] > value[e.to]){
                        value[e.to] = value[u] + weight[e.to];
                    }
                }
            }
        }
    }
}dij;
int main(){
    int n, m, st, et;
    scanf("%d%d%d%d", &n, &m, &st, &et);
    for(int i = 0; i < n; ++i){
        scanf("%d", &weight[i]);
    }
    dij.init(n);
    int x, y;
    LL l;
    for(int i = 0; i < m; ++i){
        scanf("%d%d%lld", &x, &y, &l);
        dij.AddEdge(x, y, l);
        dij.AddEdge(y, x, l);
    }
    dij.dijkstra(st);
    printf("%d %d\n", dij.num[et], dij.value[et]);
    return 0;
}