1003 Emergency (25 point(s))

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1003 Emergency (25 point(s))

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.

Example:

#include<iostream>
#include<vector>
#include<list>

using namespace std;

typedef int Vertex;

struct Link {
    Vertex vertex;
    int dist;
};
struct Graph {
    int Nv;
    int Ne;
    vector<list<Link>> Adj;
    vector<Vertex> Team;
};
struct PathType {
    int dist;
    int team;
} Path;

int Count, Dist, Teams;

void DFS(Graph &G, Vertex v, Vertex c, vector<bool> &Visited)
{
    if(v == c) {
        Path.team += G.Team[c];
        if(Dist == -1 || Path.dist < Dist) {
            Dist = Path.dist;
            Count = 1;
            Teams = Path.team;
        } else if(Path.dist == Dist) {
            Count++;
            if(Teams < Path.team)
                Teams = Path.team;
        }
        Path.team -= G.Team[c];
    } else {
        if(Visited[v]) return ;
        Visited[v] = true;
        Path.team += G.Team[v];
        for(auto &x : G.Adj[v]) {
            if(!Visited[x.vertex]) {
                Path.dist += x.dist;
                DFS(G, x.vertex, c, Visited);
                Path.dist -= x.dist;
            }
        }
        Visited[v] = false;
        Path.team -= G.Team[v];
    }
}

void ShortestPath(Graph &G, Vertex v, Vertex c)
{
    vector<bool> Visited(G.Nv, false);
    Count = 0;
    Teams = 0;
    Dist = -1;
    Path.team = 0;
    Path.dist = 0;
    DFS(G, v, c, Visited);
    cout << Count << ' ' << Teams << endl;
}

int main()
{
    int N, M, C1, C2;
    cin >> N >> M >> C1 >> C2;
    Graph G;
    G.Nv = N;
    G.Ne = M;
    G.Adj.resize(N);
    G.Team.resize(N);
    for(int i = 0; i < N; i++){
        cin >> G.Team[i];
    }
    for(int i = 0; i < M; i++){
        int c1, c2, dist;
        cin >> c1 >> c2 >> dist;
        G.Adj[c1].push_back({c2, dist});
        G.Adj[c2].push_back({c1, dist});
    }
    ShortestPath(G, C1, C2);
}

思路：

查找图的最短路径，利用DFS，然后记录下最短路径下支持的最大救援队伍数量。

本文标签： EmergencyPoint