PAT (Advanced Level) Practice 1003 Emergency (25分)

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本文介绍了一种基于图论的最短路径算法在应急救援场景中的应用，通过寻找从当前城市到目标城市的最短路径，并在此过程中尽可能多地集结救援队伍。输入包含城市数量、道路连接及救援队伍分布等信息，输出为不同最短路径的数量和最大可集结的救援队伍数量。

PAT (Advanced Level) Practice 1003 Emergency (25分)

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input Specification:

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (≤500) - the number of cities (and the cities are numbered from 0 to N−1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers C1, C2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather. All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input:

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output:

2 4

#include <bits/stdc++.h>
using namespace std;

int n, m;
vector<pair<int, int>> graph[505];
int d[505], vis[505], roadnum[505];
long long maxhands[505],a[505];
int C1, C2;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m >> C1 >> C2;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < m; i++){
        int u, v, w;
        cin >> u >> v >> w;
        graph[u].push_back({v, w});
        graph[v].push_back({u, w});
    }

    fill(d, d + n, 1e8);
    d[C1] = 0;
    maxhands[C1] = a[C1];
    roadnum[C1] = 1;

    for (int i = 0; i < n; i++){
        int minest = 1e8, minid = 0;
        for (int j = 0; j < n; j++){
            if (!vis[j] && d[j] < minest){
                minest = d[j];
                minid = j;
            }
        }
        vis[minid] = 1;
        for (auto _ : graph[minid]){
            int v = _.first;
            int w = _.second;
            if (vis[v]) continue;
            if (d[v] == d[minid] + w){
                roadnum[v] += roadnum[minid];
                maxhands[v] = max(maxhands[v], maxhands[minid] + a[v]);
            }
            else if (d[v] > d[minid] + w){
                d[v] = d[minid] + w;
                roadnum[v] = roadnum[minid];
                maxhands[v] = maxhands[minid] + a[v];
            }
        }
    }
    cout << roadnum[C2] << " " << maxhands[C2] << "\n";
    return 0;
}