1003 Emergency (25分)

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本文介绍了一个基于Dijkstra算法的紧急救援路径寻找问题。在给定的城市地图中，包含多个城市和连接它们的道路，以及每个城市的救援队伍数量。目标是在接收到其他城市的紧急呼叫时，快速找到从当前位置到目标城市的最短路径，并沿途集结尽可能多的救援队伍。

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:

Sample Output:

2 4

PS:dijksta模板（熟记），加上两个数组num和w同时进行处理就可以得出答案（需要注意是无向图）。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1000;
const int inf = 0x3fffffff;
int weight[maxn];
int map[maxn][maxn];
bool vis[maxn] = { false };
int dis[maxn],w[maxn],num[maxn];
int n, m, C1, C2;
int c1, c2, L;
void dijkstra(int s)
{
	fill(dis, dis + maxn, inf);
	memset(w, 0, sizeof(w));
	memset(num, 0, sizeof(num));
	num[s] = 1;
	w[s] = weight[s];
	dis[s] = 0;
	for (int i = 0; i < n; i++)
	{
		int u=-1, min=inf;
		for (int j = 0; j < n; j++)
		{
			if (vis[j] == false && dis[j] < min)
			{
				u = j;
				min = dis[j];
			}
		}
		if (u == -1) break;
		vis[u] = true;
		for (int v = 0; v < n; v++)
		{
			if (map[u][v] != inf && vis[v] == false)
			{
				if (dis[u] + map[u][v] < dis[v])
				{
					dis[v] = dis[u] + map[u][v];
					w[v] = weight[v] + w[u];
					num[v] = num[u];
				}
				else if (dis[u] + map[u][v] == dis[v])
				{
					if (w[v] < weight[v] + w[u])
					{
						w[v] = weight[v] + w[u];
					}
					num[v] += num[u]; 
				}
			}
		}
	}
}
int main(void)
{
	cin >> n >> m >> C1 >> C2;
	for (int i = 0; i < n; i++)
	{
		cin >> weight[i];
	}
	fill(map[0], map[0] + maxn * maxn, inf);
	for (int i = 0; i < m; i++)
	{
		cin >> c1 >> c2 >> L;
		map[c1][c2] = map[c2][c1] = L;
	}
	dijkstra(C1);
	printf("%d %d\n", num[C2],w[C2]);
	return 0;
}