题目描述

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, 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 Specification:

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:

Sample Output:

1
2
3

2 4

结尾无空行

参考柳神的题解

#include <iostream>
//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
#include <algorithm>

using namespace std;
int n, m, c1, c2;
int e[510][510],//边长
weight[510], // 城市救援队的数量
dis[510], // 最短路径
num[510], // 从出发点到当前节点i的最短路径的条数
w[510]; // 从出发点到当前节点i的 最大weight
bool visit[510];
const int inf = 99999999;

int main() {
    scanf("%d%d%d%d", &n, &m, &c1, &c2);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(visit, visit + 510, false);
    fill(dis, dis + 510, inf);
    // 读取weight
    for (int i = 0; i < n; ++i) {
        scanf("%d", &weight[i]);
    }
    // 读取每条边
    int a, b, c;
    for (int i = 0; i < m; ++i) {
        scanf("%d%d%d", &a, &b, &c);
        e[a][b] = e[b][a] = c;
    }
    //初始化
    num[c1] = 1;
    w[c1] = weight[c1];
    dis[c1] = 0;
    for (int i = 0; i < n; ++i) {
        int temp = -1;
        int min = inf;
        // 拿到dist中最小的节点，可以用最小索引堆进行优化
        for (int j = 0; j < n; ++j) {
            if (!visit[j] && dis[j] < min) {
                temp = j;
                min = dis[j];
            }
        }
        if (temp == -1) break; //说明输入有误
        visit[temp] = true;
        // 遍历每个顶点的周围顶点
        for (int v = 0; v < n; ++v) {
            if (!visit[v] && e[temp][v] != inf) {
                // 如果找到比当前最短路径还要短的路径，就更新dist
                if (dis[temp] + e[temp][v] < dis[v]) {
                    dis[v] = dis[temp] + e[temp][v];
                    num[v] = num[temp];
                    w[v] = w[temp] + weight[v];
                } else if (dis[temp] + e[temp][v] == dis[v]) {
                    // 找到和dist[v] 一样长度的，说明有另外一条最短路径，更新num[v]
                    num[v] = num[temp] + num[v];
                    // 根据题目，如果最短路径 >=2 就需要开始比较 w[v] ，也就是比较这>=2 条的最短路径中，谁的weight之和 最大
                    // 从而找到最多的救援队
                    if(w[temp] +weight[v] > w[v]){
                        w[v] = w[temp] + weight[v];
                    }
                }
            }
        }
    }

    cout << num[c2] << " " << w[c2] << endl;

    return 0;
}