PAT 1003 Emergency (Dijkstra算法 ) 示例解析

程序员文章站 2022-06-11 11:16:54

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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:

Sample Output:

2 4

题目含义：输入城市数目n个，城市道路m条，每个城市有自己的救援小组，所有的路的权重已知。给定起点和终点，求出起点到重点的最短路径数目以及最短路径上的救援小组数目之和。

测试网址：https://pintia.cn/problem-sets/994805342720868352/problems/994805523835109376

分析：使用Dijkstra算法

救援小组个数相当于点权，用Dijkstra求出边全最小的最短路径的条数，以及这些最短路径中最大点权值。dis[i] 表示从出发点到i节点最短路径的路径长度,roads[i] 表示从出发点到i节点最短路径的条数，teams[i] 表示从出发点到i点救援队的数目之和，e[u][v] 表示u城市到v城市的路径权重

伪代码：
if dis[u] + e[u][v] < dis[v] 
    dis[v] = dis[u] + e[u][v];
    roads[v] = roads[u];
    teams[v] = teams[u] + weight[v]; 
else if dis[u] + e[u][v] == dis[v]
    roads[v] += roads[u];
    if teams[u] + weight[v] > teams[v]
         teams[v] = teams[u] + weight[v];

实际代码：

#include <iostream>
#include <vector>
#include <limits.h>

using namespace std;

int main() {  
    //读取第一行数据
    int n, m, C1, C2;
    cin>>n>>m>>C1>>C2;
    vector<int> weight(n, 0);
    for (int i = 0; i < n; i++) {
        cin>>weight[i];
    }
    
    vector<vector<int> > edge(n, vector<int>(n, INT_MAX));
    vector<int> dis(n, INT_MAX);
    vector<int> roads(n, 0);
    vector<int> teams(n, 0);
    vector<bool> visit(n, false);

    //初始化边权值表
    int c1, c2, L;
    for (int i = 0; i < m; i++) {
        cin>>c1>>c2>>L;
        edge[c1][c2] = edge[c2][c1] = L;
    }
    dis[C1] = 0;
    teams[C1] = weight[C1];
    roads[C1] = 1;
  
    for (int i = 0; i < n; ++i) {
        int u = -1, minn = INT_MAX;
        for (int j = 0; j < n; j++) {
            if (visit[j] == false && dis[j] < minn) {
                u = j;
                minn = dis[j];
            }
        }
        if (u == -1) break;
        visit[u] = true;
        for (int v = 0; v < n; v++) {
            if (visit[v] == false && edge[u][v] != INT_MAX) {
                if (dis[u] + edge[u][v] < dis[v]) {
                    dis[v] = dis[u] + edge[u][v];
                    roads[v] = roads[u];
                    teams[v] = teams[u] + weight[v];
                } else if (dis[u] + edge[u][v] == dis[v]) {
                    roads[v] = roads[v] + roads[u];
                    if (teams[u] + weight[v] > teams[v]) {
                        teams[v] = teams[u] + weight[v];
                    }
                }
            }
        }
    }
    cout<<roads[C2]<<" "<<teams[C2]<<endl;
    return 0;
}

示例解析：

输入参数

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