path

path

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1100    Accepted Submission(s): 219

Problem Description

You have a directed weighted graph with n vertexes and m edges. The value of a path is the sum of the weight of the edges you passed. Note that you can pass any edge any times and every time you pass it you will gain the weight.

Now there are q queries that you need to answer. Each of the queries is about the k-th minimum value of all the paths.

 

 

Input

The input consists of multiple test cases, starting with an integer t (1≤t≤100), denoting the number of the test cases.
The first line of each test case contains three positive integers n,m,q. (1≤n,m,q≤5∗104)

Each of the next m lines contains three integers ui,vi,wi, indicating that the i−th edge is from ui to vi and weighted wi.(1≤ui,vi≤n,1≤wi≤109)

Each of the next q lines contains one integer k as mentioned above.(1≤k≤5∗104)

It's guaranteed that Σn ,Σm, Σq,Σmax(k)≤2.5∗105 and max(k) won't exceed the number of paths in the graph.

 

 

Output

For each query, print one integer indicates the answer in line.

 

 

Sample Input

1 2 2 2 1 2 1 2 1 2 3 4

 

 

Sample Output

3 3

Hint

1->2 value :1 2->1 value: 2 1-> 2-> 1 value: 3 2-> 1-> 2 value: 3

 

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 1e5 + 10;
struct node{
    ll to,val;
    bool operator<(const node& s)const {
        return val>s.val;
    }
};
int T,n,m,q,query[maxn];
ll ans[maxn];
vector<pii>e[maxn];
priority_queue<node>Q;
multiset<int>s;
int main() {
    //freopen("1.txt", "r", stdin);
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d",&n,&m,&q);
        s.clear();
        while(!Q.empty())Q.pop();
        for(register int i=1;i<=n;++i)e[i].clear();
        for(register int i=1,u,v,w;i<=m;++i){
            scanf("%d%d%d",&u,&v,&w);
            e[u].emplace_back(make_pair(w,v));
            s.insert(w);
            Q.push(node{v,w});
        }
        for(register int i=1;i<=n;++i){
            sort(e[i].begin(),e[i].end());
        }
        int mx=-1;
        for(register int i=1,k;i<=q;++i){
            scanf("%d",&query[i]);
            mx=max(mx,query[i]);
        }
        int cnt=0;
        while(cnt<mx){
            node cur=Q.top();
            Q.pop();
            ans[++cnt]=cur.val;
            if(cnt>=mx)break;
            int u=cur.to;
            for(auto it:e[u]){
                ll v=it.second;
                ll d=cur.val+it.first;
                if(s.size()==mx){
                    auto po=--s.end();
                    if(d>=*po)break;
                    s.erase(po);
                    s.insert(d);
                }
                else{
                    s.insert(d);
                }
                Q.push(node{v,d});
            }
        }
        for(register int i=1;i<=q;++i){
            printf("%d\n",ans[query[i]]);
        }
    }
    return 0;
}