Portal 的问题

Question

Problem Description
ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.

Input
There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).

Output
Output the answer to each query on a separate line.

Sample Input
10 10 10
7 2 1
6 8 3
4 5 8
5 8 2
2 8 9
6 4 5
2 1 5
8 10 5
7 3 7
7 8 8
10
6
1
5
9
1
8
2
7
6

Sample Output
36
13
1
13
36
1
36
2
16
13

Answer 1

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n,m,q,l;
vector<int>adj[maxn];
int dis[maxn],f[maxn];
void dijkstra(int s){
    priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> q;
    memset(dis,-1,sizeof(dis));
    dis[s] = 0;
    q.push(make_pair(0,s));
    while(!q.empty()){
        auto t = q.top();
        q.pop();
        if(t.first > dis[t.second]) continue;
        for(auto &u: adj[t.second]){
            if(dis[u] <= -t.first + f[u]) continue;
            dis[u] = -t.first + f[u];
            q.push(make_pair(-dis[u],u));
        }
    }
}
int main(){
    cin >> n >> m >> q;
    l = 0;
    for(int i=1;i<=m;++i){
        int u,v,c;
        cin >> u >> v >> c;
        adj[u].push_back(v);
        adj[v].push_back(u);
        ++l;
    }
    for(int i=1;i<=n;++i){
        for(int j=0;j<adj[i].size();++j){
            int u = adj[i][j];
            f[u] += c;
        }
    }
    for(int i=1;i<=n;++i){
        if(l == 1 && i != 1) break;
        dijkstra(i);
        cout << dis[n] << endl;
    }
    return 0;
}