HDU-6705 Path

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. (\(1≤n,m,q≤5∗10^4\))

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.(\(1≤k≤5∗10^4\))

It's guaranteed that \(Σn ,Σm, Σq,Σmax(k)≤2.5∗10^5\) 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

题解

给定一张有向图，q次询问，每次询问第k小的路径长度。

离线，预处理出最大的k范围内的全部路径长度。先将全部边按边权排序，用一个set存储当前能够成为答案的边，且set的最大的大小为maxk，每次从set中取出w最小的边，看看可否更新set中的元素，不能更新则break（边权从小到大排序，小边权没法更新以后边权也没法更新），对set中的元素都作一次这样的处理后，咱们就获得了[1,maxk]的答案，输出询问便可，复杂度\(O(k*log(m+k))\)

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e4 + 50;
struct node {
    int v; ll w;
    node (int v = 0, int w = 0): v(v), w(w) {}
    bool operator < (const node &b) const {
        return w < b.w;
    }
};
vector<node> G[N];
struct Edge {
    int u, v; ll w;
    int id;
    Edge(int u = 0, int v = 0, ll w = 0, int id = 0): u(u), v(v), w(w), id(id) {}
    bool operator < (const Edge &b) const {
        if (w == b.w)
            if (u == b.u)
                if (v == b.v)
                    return id < b.id;
                else return v < b.v;
            else return u < b.u;
        else return w < b.w;
    }
    bool operator == (const Edge &b) const {
        return w == b.w && u == b.u && v == b.v && id == b.id;
    }
};
int Q[N];
ll ans[N];
int main() {
    int t; scanf("%d", &t);
    while (t--) {
        int n, m, q;
        scanf("%d%d%d", &n, &m, &q);
        for (int i = 1; i <= n; i++) G[i].clear();
        set<Edge> st; st.clear();
        int cnt = 0;
        for (int i = 1; i <= m; i++) {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            G[u].push_back(node(v, w));
            st.insert(Edge(u, v, w, ++cnt));
        }
        for (int i = 1; i <= n; i++) sort(G[i].begin(), G[i].end());
        int maxk = 0;
        for (int i = 1; i <= q; i++) {
            scanf("%d", &Q[i]);
            maxk = max(maxk, Q[i]);
        }
        while (st.size() > maxk) st.erase(st.end());
        for (int i = 1; i <= maxk; i++) {
            Edge now = *st.begin();
            st.erase(st.begin());
            ans[i] = now.w;
            if (i == maxk) break;
            int u = now.v;
            for (int j = 0; j < G[u].size(); j++) {
                int v = G[u][j].v;
                ll w = G[u][j].w;
                if (i + st.size() < maxk) st.insert(Edge(now.u, v, now.w + w, ++cnt));
                else {
                    set<Edge>::iterator it = st.end(); it--;
                    Edge last = *it;
                    if (now.w + w < last.w) {
                        st.erase(it);
                        st.insert(Edge(u, v, now.w + w, ++cnt));
                    }
                    else break;
                }
            }
        }
        for (int i = 1; i <= q; i++) printf("%lld\n", ans[Q[i]]);
    }
    return 0;
}