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

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

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
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

翻译:

作为一个城市紧急营救队队长,你有一份特殊的地图,地图上有几个城市,它们以道路相连,每个城市的营救队伍和分别与其他城市连接的道路长度都在地图上有标注,当某个城市发生紧急事件时,你的任务是尽快带领你们队员到达该城市,与此同时,尽可能带上更多的队员。

输入:

每一个输入文件包含一个测试用例,对于每一个用例,第一行包含4个整数:N(<=500)——城市的总数(从0到 N-1),M——道路的条数,C1和C2——分别为你所在的城市和你要营救的城市,下一行包含N个整数,分别表示N个城市的各个营救队员数,随后有M行,每一行有C1,C2,L,表示每个城市之间的道路长度,保证C1和C2至少存在一条路。

输出:

对于每一个测试用例,在一行内输出两个数字:C1和C2之间不同最短路径的数量,以及你能收集到的最大救援数。

思路:

最短路径问题,用dijkstra算法,道路的边权已知,求从起点到终点的最短路径以及最短路径的条数,如有多条最短路径,输出点权(救援队员数量)最大的那条。

dis[i]:起点到i的最短距离。

num[i]:从起点到i最短路径的条数。

totalWtg[i]:从起点到i的点权最大的最短路径的权重数。


#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
using namespace std;
int n, m, c1, c2;
int e[510][510], cityWtg[510], dis[510], num[510], totalWtg[510];
bool visit[510];
const int inf = 99999999;
int main() {
	scanf("%d%d%d%d", &n, &m, &c1, &c2);
	for (int i = 0; i < n; i++)
		scanf("%d", &cityWtg[i]);
	fill(e[0], e[0] + 510 * 510, inf);
	fill(dis, dis + 510, inf);
	int a, b, c;
	for (int i = 0; i < m; i++) 
	{
		scanf("%d%d%d", &a, &b, &c);
		e[a][b] = c;
		e[b][a] = c;
	}
	dis[c1] = 0;
	totalWtg[c1] = cityWtg[c1];
	num[c1] = 1;
	for (int i = 0; i < n; i++) 
	{
		int u = -1, minn = inf;
		for (int j = 0; j < n; j++) 
		{
			if (visit[j] == false && dis[j] < minn) //没有visit过,且到起点的值最小,把u设为下一个current点
			{	
				u = j;
				minn = dis[j];
			}
		}
		if (u == -1) break;
		visit[u] = true;
		for (int v = 0; v < n; v++) 
		{
			if (visit[v] == false && e[u][v] != inf) 
			{ 
				if (dis[u] + e[u][v] < dis[v])//当u的邻点+u到领点的距离小于之前路线保存的值 
				{ // 替换为新的最短路线  
					dis[v] = dis[u] + e[u][v]; 
					num[v] = num[u]; 
					totalWtg[v] = totalWtg[u] + cityWtg[v]; 
				} 
				else if (dis[u] + e[u][v] == dis[v]) 
				{ 
					num[v] = num[v] + num[u]; 
					if (totalWtg[u] + cityWtg[v] > totalWtg[v])// 如果点权大,取大点权 
					totalWtg[v] = totalWtg[u] + cityWtg[v];
				}
			}
		}
	}
	printf("%d %d", num[c2], totalWtg[c2]);
	return 0;
}

如果一下子不能理解,可以举个例子画个图,思路跟着程序单步走一遍。另, Dijkstra算法详解