1003. Emergency (25)

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.

5 6 0 21 2 1 5 30 1 10 2 20 3 11 2 12 4 13 4 1Sample Output
2 4
#include<cstdio>#include<vector>#include<queue>#include<algorithm>using namespace std;const int maxn = 500;const int INF = 1000000000;struct Node{	int v, dis;}node;struct compare{	bool Operator()(Node n1, Node n2)	{		return n1.dis > n2.dis;	}};//自定義比較器vector<Node> Adj[maxn];//圖的鄰接表形式int N, M, C1, C2, c1, c2, L;int weight[maxn] = { 0 };//記錄各頂點的權(quán)bool vis[maxn] = { false };//標(biāo)記頂點是否被訪問int d[maxn], w[maxn],//統(tǒng)計最短路徑、最大點權(quán)num[maxn];//統(tǒng)計最短路徑數(shù)void Dijkstra(int s){	fill(d, d + maxn, INF);	fill(w, w + maxn, 0);	fill(num, num + maxn, 0);	d[s] = 0;	w[s] = weight[s];//對增加點權(quán)的問題初始化	num[s] = 1;//對增加路徑數(shù)的問題初始化	priority_queue<Node,vector<Node>,compare>Q;//找出最小的u,我這里沒用窮舉方式，用堆優(yōu)化一下	node.v = s; node.dis = d[s];	Q.push(node);	int u;	for (int i = 0; i < N; i++)	{		if (!Q.empty())		{			u = Q.top().v;			vis[u] = true;			Q.pop();		}		else			return;//已經(jīng)沒有要處理的點了，可以返回了		for (int j = 0; j < Adj[u].size(); j++)		{			int v = Adj[u][j].v;			int dis = Adj[u][j].dis;//注意這里是邊長			if (!vis[v])			{				if (d[u] + dis < d[v])				{					d[v] = d[u] + dis;					w[v] = weight[v] + w[u];					num[v] = num[u];//優(yōu)化情形下直接繼承					node.v = v; node.dis = d[v];//注意這里的dis是更新為起點到當(dāng)前點的距離					Q.push(node);				}				else if (d[u] + dis == d[v])				{					num[v] += num[u];//相等則累加					if(w[v]<weight[v] + w[u])						w[v] = weight[v] + w[u];//總是更新為最大點權(quán)                                                                                                    				}			}		}	}}int main(){	scanf("%d%d%d%d", &N, &M, &C1, &C2);	for (int i = 0; i < N; i++)	{		scanf("%d", &weight[i]);	}	for (int i = 0; i < M; i++)	{		scanf("%d%d%d", &c1, &c2, &L);		node.v = c2;		node.dis = L;		Adj[c1].push_back(node);		node.v = c1;		Adj[c2].push_back(node);	}	Dijkstra(C1);	printf("%d %d/n", num[C2], w[C2]);	return 0;}