pat a1003 Emergency 【dijkstra求最短路径】

题目：

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

一、问题描述

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, C​1​​ and C​2​​ - 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 c​1​​, c​2​​ 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 C​1​​ to C​2​​.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between C​1​​ and C​2​​, 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

 给出N个城市，M条无向边。每个城市中都有一定数目的救援小组，所有边的边权已知。现在给出起点和终点，求从起点到终点的最短路径条数及最短路径上的救援小组数目之和。如果有多条最短路径，则输出数目之和最大的。

二、算法实现

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxv=510;
const int inf=1000000000;

int n,m;  //顶点数，边数 
int st,ed;  //起点，终点
int G[maxv][maxv],weight[maxv];  //邻接矩阵，每个点的权值 

//从起点到u 
int d[maxv];  //最短距离  
int w[maxv];  //最大点权之和
int num[maxv];  //最短路径条数

bool vis[maxv]={false};

void Dijkstra(int s) 
{
	fill(d,d+maxv,inf);
	memset(num,0,sizeof(num));
	memset(w,0,sizeof(w));
	d[s]=0;
	w[s]=weight[s];
	num[s]=1;
	for(int i=0;i<n;i++)
	{
		int u=-1;
		int min=inf;
		for(int j=0;j<n;j++)
		{
			if(vis[j]==false&&d[j]<min)
			{
				u=j;
				min=d[j];
			}
		}
		if(u==-1) return;
		vis[u]=true;
		for(int v=0;v<n;v++)
		{
			if(vis[v]==false&&G[u][v]!=inf)
			{
				if(d[u]+G[u][v]<d[v])
				{
					d[v]=d[u]+G[u][v];
					w[v]=w[u]+weight[v];
					num[v]=num[u];
				}
				else if(d[u]+G[u][v]==d[v])
				{
					if(w[u]+weight[v]>w[v])
					{
						w[v]=w[u]+weight[v];
					}
					num[v]+=num[u];
				}
			}
		}
	}
}
 
int main()
{
	scanf("%d%d%d%d",&n,&m,&st,&ed);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&weight[i]);
	}
		
	fill(G[0],G[0]+maxv*maxv,inf);
	
	int u,v;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&u,&v);
		scanf("%d",&G[u][v]);
		G[v][u]=G[u][v]; 
	}
	Dijkstra(st);
	printf("%d %d\n",num[ed],w[ed]);
	return 0; 
}