NC24755 [USACO 2010 Dec S]Apple Delivery

题目链接

题目

题目描述

Bessie has two crisp red apples to deliver to two of her friends in the herd. Of course, she travels the C (1 <= C <= 200,000)
cowpaths which are arranged as the usual graph which connects P (1 <= P <= 100,000) pastures conveniently numbered from 1..P: no cowpath leads from a pasture to itself, cowpaths are bidirectional, each cowpath has an associated distance, and, best of all, it is always possible to get from any pasture to any other pasture. Each cowpath connects two differing pastures $P1_i$ (1 <= $P1_i$ <= P) and $P2_i$​ (1 <= $P2_i$ <= P) with a distance between them of $D_i$. The sum of all the distances $D_i$​ does not exceed 2,000,000,000.
What is the minimum total distance Bessie must travel to deliver both apples by starting at pasture PB (1 <= PB <= P) and visiting pastures PA1 (1 <= PA1 <= P) and PA2 (1 <= PA2 <= P)
in any order. All three of these pastures are distinct, of course.

Consider this map of bracketed pasture numbers and cowpaths with distances:
                3        2       2
           [1]-----[2]------[3]-----[4]
             \     / \              /
             7\   /4  \3           /2
               \ /     \          /
               [5]-----[6]------[7]
                    1       2
If Bessie starts at pasture [5] and delivers apples to pastures [1] and [4], her best path is:
      5 -> 6-> 7 -> 4* -> 3 -> 2 -> 1*
with a total distance of 12.

输入描述

• Line 1: Line 1 contains five space-separated integers: C, P, PB, PA1, and PA2
• Lines 2..C+1: Line i+1 describes cowpath i by naming two pastures it connects and the distance between them: $P1_i, P2_i, D_i$

输出描述

• Line 1: The shortest distance Bessie must travel to deliver both apples

示例1

输入

9 7 5 1 4 
5 1 7 
6 7 2 
4 7 2 
5 6 1 
5 2 4 
4 3 2 
1 2 3 
3 2 2 
2 6 3 

输出

12

题解

知识点：最短路。

从 $PB$ 出发必须经过 $PA1,PA2$ 的最短路，显然我们分别考虑 $PA1,PA2$ 作为终点即可。

先以三个点为起点跑三次最短路，然后讨论两条路径 PB->PA1->PA2,PB->PA2->PA1 的最小值即可。

时间复杂度 $O((C+P)\log C)$

空间复杂度 $O(C+P)$

代码

#include <bits/stdc++.h>
#define ll long long
 
using namespace std;
 
const int N = 100007, M = 200007 << 1;
 
template<class T>
struct Graph {
    struct edge {
        int v, nxt;
        T w;
    };
    int idx;
    vector<int> h;
    vector<edge> e;
 
    Graph(int n, int m) :idx(0), h(n + 1), e(m + 1) {}
 
    void clear(int n, int m) {//全局使用时清零，确定范围防止超时
        idx = 0;
        h.assign(n + 1, 0);
        e.assign(m + 1, { 0,0,0 });
    }
 
    void add(int u, int v, T w) {
        e[++idx] = edge{ v,h[u],w };
        h[u] = idx;
    }
};
Graph<int> g(N, M);
 
void dijkstra(int st, vector<int> &dis) {
    dis.assign(dis.size(), 0x3f3f3f3f);
    vector<bool> vis(dis.size(), 0);
    struct node {
        int v, w;
        bool operator<(node a) const {
            return w > a.w;
        }
    };
    priority_queue<node> pq;
    dis[st] = 0;
    pq.push({ st,0 });
    while (!pq.empty()) {
        int u = pq.top().v;
        pq.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        for (int i = g.h[u];i;i = g.e[i].nxt) {
            int v = g.e[i].v, w = g.e[i].w;
            if (dis[v] > dis[u] + w) {
                dis[v] = dis[u] + w;
                pq.push({ v, dis[v] });
            }
        }
    }
}
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int C, P, PB, PA1, PA2;
    cin >> C >> P >> PB >> PA1 >> PA2;
    for (int i = 1;i <= C;i++) {
        int P1, P2, D;
        cin >> P1 >> P2 >> D;
        g.add(P1, P2, D);
        g.add(P2, P1, D);
    }
    vector<int> disB(P + 1), disA1(P + 1), disA2(P + 1);
    dijkstra(PB, disB);
    dijkstra(PA1, disA1);
    dijkstra(PA2, disA2);
    cout << min(disB[PA1] + disA1[PA2], disB[PA2] + disA2[PA1]) << '\n';
    return 0;
}