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, 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

一，注意事項：

    使用Dijsktra演算法之前一定要注意初始化，題目中有什麼指標就對什麼初始化，所有的都要先對距離（或花費cost[max_n]等）dis[max_n]等初始化，還要對第二指標路徑的數目num[max_n]，還有救援人員的人數amount[start]=single[start]，當初就忘了對這個進行初始化。

#include<cstdio>
#include<algorithm>
using namespace std;
const int INF = 1000000000;
const int max_n = 550;
int N = 0;
int G[max_n][max_n] = { 0 };
int dis[max_n] = { 0 };
int num[max_n] = { 0 };
int amount[max_n] = { 0 };
int single[max_n] = { 0 };
bool vis[max_n] = { false };
void Dijkstra(int start) {
	fill(dis, dis + max_n, INF);
	dis[start] = 0;
	num[start] = 1;
	amount[start] = single[start];
	for (int i = 0; i < N; i++) {
		int MIN = INF, u = -1;
		for (int j = 0; j < N; j++) {
			if (dis[j] < MIN&&vis[j] == false) {
				u = j;
				MIN = dis[j];
			}
		}
		if (u == -1)return;
		vis[u] = true;
		for (int j = 0; j < N; j++) {
			if (vis[j] == false && G[u][j] != 0) {
				if (dis[u] + G[u][j] < dis[j]) {
					dis[j] = dis[u] + G[u][j];
					amount[j] = amount[u] + single[j];
					num[j] = num[u];
				}
				else if (dis[u] + G[u][j] == dis[j]) {
					num[j] += num[u];
					if (amount[u] + single[j] > amount[j]) {
						amount[j] = amount[u] + single[j];
					}
				}
			}
		}
	}
}
int main() {
	int M = 0;
	int c1 = 0, c2 = 0;
	int x = 0, y = 0, z = 0;
	scanf("%d %d %d %d", &N, &M, &c1, &c2);
	for (int i = 0; i < N; i++) {
		scanf("%d", &single[i]);
	}
	for (int i = 0; i < M; i++) {
		scanf("%d %d %d", &x, &y, &z);
		G[x][y] = z;
		G[y][x] = z;
	}
	Dijkstra(c1);
	printf("%d %d", num[c2], amount[c2]);
	return 0;
}