2019ccpc网络赛hdu6705 path

path

题目传送门

解题思路

先用vector存图，然后将每个vector按照边的权值从小到大排序。将每个顶点作为起点的边里最短的边存入优先队列。对于存入优先队列的信息，应有边起点，终点，是其起点的第几短边，以及路径总长度。优先队列按照路径长度从小到大排序，每次出队当前最短的路径，对于一条路径，更新两条新的可能最短的路径，即这条路后面加上一条可走的最短边，以及这条路最后一条边换成一条次短边。将询问排序，不断更新答案即可。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 100010;
struct T{
    int x, y, f;
    ll w;
    T(int x, int y, ll w, int f): x(x), y(y), w(w), f(f){}
    bool operator<(const T& a)const{
        return w > a.w;
    }
};
struct line{
    int r;
    ll w;
    line(int r, ll w): r(r), w(w){}
    bool operator<(const line& a)const{
        return w < a.w;
    }
};
vector<line> vec[N];
struct R{
    int i, k;
    bool operator<(const R& a)const{
        return k < a.k;
    }
}qy[N];
ll ans[N];
priority_queue<T> pq;
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 <= m; i ++){
            int u, v;
            ll w;
            scanf("%d%d%lld", &u, &v, &w);
            vec[u].push_back(line(v, w));
        }
        for(int i = 1; i <= n; i ++){
            sort(vec[i].begin(), vec[i].end());
        }
        for(int i = 1; i <= n; i ++){
            if(vec[i].size())
                pq.push(T(i, vec[i][0].r, vec[i][0].w, 0));
        }
        for(int i = 1; i <= q; i ++){
            scanf("%d", &qy[i].k);
            qy[i].i = i;
        }
        sort(qy + 1, qy + q + 1);
        int cnt = 0;
        int id = 1;
        while(!pq.empty()){
            int x = pq.top().x;
            int y = pq.top().y;
            ll w = pq.top().w;
            int f = pq.top().f;
            pq.pop();
            ++cnt;
            while(id <= q && cnt == qy[id].k){
                ans[qy[id].i] = w;
                ++id;
            }
            if(id > q)
                break;
            if(vec[y].size())
                pq.push(T(y, vec[y][0].r, w + vec[y][0].w, 0));
            if(vec[x].size() > f + 1)
                pq.push(T(x, vec[x][f + 1].r, w + vec[x][f + 1].w - vec[x][f].w, f + 1));
        }
        for(int i = 1; i <= q; i ++){
            printf("%lld\n", ans[i]);
        }
        for(int i = 0; i <= n; i ++)
            vec[i].clear();
        while(!pq.empty())
            pq.pop();
    }
    return 0;
}