【题解】【PAT甲】1003 Emergency (25 分)（Dijkstra）（最短路径数量）

本文链接：https://blog.youkuaiyun.com/Elephant_King/article/details/121852216

该博客介绍了如何使用Dijkstra算法解决一个紧急救援场景下的图论问题。给定一个城市地图，包含多个城市和道路，以及每个城市的救援队伍数量，任务是从当前城市出发，找到到达目标城市的最短路径，并在此过程中尽可能多地召集救援队伍。博主提供了详细的解题思路和代码实现，强调在计算最短路径条数时不能使用INT_MAX，而应该用一个较大的常数表示无穷。

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题目链接

PTA | 程序设计类实验辅助教学平台

题目描述

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:

结尾无空行

Sample Output:

2 4

结尾无空行

题目大意

给一个图的起点终点还有每个节点的人数，让你求从起点到终点的最短路径有几条，在选择一条最短的而且总人数最多的

解题思路

Dijkstra

求最短路径的条数是设一个disnum数组

如果最短路径更新就

disnum[v]=disnum[u];

如果最短路径相同就

disnum[v]=disnum[u]+disnum[v];

相当于对于v节点，每多一个v，就相当于前u个都走一遍新多出来的v，所以是加disnum[u]

补充（2022年2月21日）：

这里取最大值不能用INT_MAX,否则取出来的值为-1

用INF=0x3f3f3f3f

题解

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int g[505][505];
int dis[505];
int people[505];
int disnum[505];
int num[505];
int visited[505];
int main(){
	int n,m,be,en;
	cin>>n>>m>>be>>en;
	memset(dis,INF,sizeof(dis));
	memset(g,INF,sizeof(g));
	for(int i=0;i<n;i++)	cin>>num[i];
	for(int i=0;i<m;i++){
		int a,b,len;
		cin>>a>>b>>len;
		g[a][b]=g[b][a]=len;
	}
	int sum=1;
	dis[be]=0;
	people[be]=num[be];
	disnum[be]=1;
	for(int i=0;i<n;i++){
		int mmin=INF;
		int u=-1;
		for(int j=0;j<n;j++){
			if(visited[j]==0&&dis[j]<mmin){
				mmin=dis[j];
				u=j;
			}
		}
		if(u==-1)	break;
		visited[u]=1;
		for(int v=0;v<n;v++){
			if(visited[v]==0){
				if(dis[v]>dis[u]+g[u][v]){
					dis[v]=dis[u]+g[u][v];
					people[v]=people[u]+num[v];
					disnum[v]=disnum[u];
				}
				else if(dis[v]==dis[u]+g[u][v]){
					disnum[v]=disnum[u]+disnum[v];
					if(people[v]<people[u]+num[v])	people[v]=people[u]+num[v];
				}	
			}
			
		}
	} 
	cout<<disnum[en]<<" "<<people[en];
}