PAT - 甲级 - 1003. Emergency (25)（Dijkstra）

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

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题目要求：
1.从a点到b点的最短路有几条？
2.在最短路中，有一条路上的所有节点的权之和最大是多少？
题目解答：
1.2.Dijkstra中在每次进行更新最短路的时候，即 dis[j] = max(dis[j], dis[x] + G[x][j]) 拆开来判断，（小于和等于的情况，在两种情况下进行相应的更新操作即可。）
具体操作看代码。

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#include<iostream>
#include<cstdio>
#include<vector>
#define INF 100000
using namespace std;
int vis[505]; //访问标记
int G[505][505]; //边权值
int dis[505]; //起点到各点距离
int num[505];	//起点到各点的条数
int w[505]; //起点到各点的队伍数
int weight[505]; //各点的队伍数
int N, M, C1, C2;

void dijkstra(){
	w[C1] = weight[C1];
	num[C1] = 1;

	for(int i = 0; i < N; i++){
		int x = -1, mdis = INF;
		for(int j = 0; j < N; j++){
			if(!vis[j] && dis[j] < mdis){
				mdis = dis[j];
				x = j;
			}
		}
		if(x==-1) break;
		vis[x] = 1;
		for(int j = 0; j < N; j++){
			if(!vis[j] && G[x][j] != INF){
				if(dis[x] + G[x][j] < dis[j]){
					dis[j] = dis[x] + G[x][j];
					num[j] = num[x];
					w[j] = w[x] + weight[j];
				}else if(dis[x] + G[x][j] == dis[j]){
					num[j] += num[x];
					if(w[x] + weight[j] > w[j]){
						w[j] = w[x] + weight[j];
					}
				}
			}

		}
	}
}
void init(){
	int a,b,c;
	while(scanf("%d%d%d%d",&N, &M, &C1, &C2) != EOF){
		// 初始化
		for(int i = 0; i < N; i++){
			dis[i] = i == C1 ? 0 : INF;
			num[i] = i == C1 ? 1 : 0;
			w[i] = 0;
			vis[i] = 0;
			for(int j = 0; j < N; j++){
				G[i][j] = INF;
			}
		}
		//input
		for(int i = 0; i < N; i++){
			scanf("%d",&weight[i]);
		}
		for(int i = 0; i < M; i++){
			scanf("%d%d%d",&a, &b, &c);
			G[a][b] = c;
			G[b][a] = c;
		}
	}
}
int main(){
//  freopen("input.txt", "r", stdin);
	init();
	dijkstra();
	cout<<num[C2]<<" "<<w[C2]<<endl;
	return 0;
}