1003. Emergency (25)-PAT甲级真题（Dijkstra算法）

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.

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
题目大意：n个城市m条路，每个城市有救援小组，所有的边的边权已知。给定起点和重点，求从起点到重点的最短路径条数以及最短路径上的救援小组数目之和。如果有多条就输出点权（城市救援小组数目）最大的那个~

分析：用一遍dijkstra算法。设立num[i]和w[i]表示从出发点到i结点拥有的路的条数，以及能够找到的救援队的数目~~~当判定dis[u] + e[u][v] < dis[v]的时候，不仅仅要更新dis[v]，还要更新num[v] = num[u], w[v] = weight[v] + w[u]; 如果dis[u] + e[u][v] == dis[v]，还要更新num[v] += num[u]，而且判断一下是否权重w[v]更小，如果更小了就更新w[v] = weight[v] + w[u];

#include <cstdio>#include <algorithm>using namespace std;int n, m, c1, c2;int dis[510], weight[510], e[510][510], num[510], w[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", &weight[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;    w[c1] = weight[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) {                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]) {                    dis[v] = dis[u] + e[u][v];                    num[v] = num[u];                    w[v] = w[u] + weight[v];                } else if(dis[u] + e[u][v] == dis[v]) {                    num[v] = num[v] + num[u];                    if(w[u] + weight[v] > w[v])                        w[v] = w[u] + weight[v];                }            }        }    }    printf("%d %d", num[c2], w[c2]);    return 0;}