1003. Emergency (25)

最新推荐文章于 2021-01-20 12:22:59 发布

转载最新推荐文章于 2021-01-20 12:22:59 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/minesweeper/p/6271365.html

本文介绍了一种基于Dijkstra算法的变种，用于解决特定问题：找出两点间的所有最短路径数量，并在这些路径中寻找能集结最多救援队伍的方案。通过调整传统Dijkstra算法，确保了效率的同时满足了问题的特殊需求。

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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

Sample Output

2 4

/* 
  题意：求最短路的总条数 并在保证最短路的情况下  求一条最短路  路径中所有城市中的护士总人数最大。
  dijsktra的变种 想明白就好写。
 */
#include "iostream"
using namespace std;
#define INF 999999
int p[501], map[501][501], dist[501];
int Count[501];
int cost[501];
int n, m, s, e;
bool flag = 0;
bool vis[501] = { false };
void dijkstra() {
    int u ;
    for (int j = 0; j < n-1; j++) {
        int Min = INF;
        for (int i = 0; i < n; i++)
        {
            if (!vis[i] && dist[i] < Min) {
                Min = dist[i];
                u = i;
            }
        }
        if (Min < INF) 
            vis[u] = true;
        else { /* 图不连通 结束*/
            flag = 1;
            break;
        }
        for (int i = 0; i < n; i++) {
            if (!vis[i]) {
                if (dist[i] >= dist[u] + map[u][i]) {
                    if (dist[i] == dist[u] + map[u][i]) {
                        Count[i] += Count[u];
                        if (cost[i] < cost[u] + p[i])
                            cost[i] = cost[u] + p[i];
                    }
                    else {
                        dist[i] = dist[u] + map[u][i];
                        Count[i] = Count[u];
                        cost[i] = cost[u] + p[i];
                    }
                }
            }
        }
    }
}
int main() {
    cin >> n >> m >> s >> e;
    for(int i=0; i<n; i++)
        for (int j = 0; j < n; j++) {
            if (i == j)
            map[i][j] = map[j][i] = 0;
            else
            map[i][j] = map[j][i] = INF;
        }
    for (int i = 0; i < n; i++) {
        cin >> p[i];
    }
    for (int i = 0; i < m; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        map[a][b] = map[b][a] = c;
    }
    for (int i = 0; i < n; i++) {
        dist[i] = map[i][s];
    }
    cost[s] = p[s];
    Count[s] = 1;
    dijkstra();
    if (flag)
        cout << 0 << " " << 0 << endl;
    else
    cout << Count[e] <<" "<< cost[e] << endl;
    return 0;
}

转载于:https://www.cnblogs.com/minesweeper/p/6271365.html