zoukankan      html  css  js  c++  java
  • PAT甲级 1003 【Emergency】

    1003 Emergency (25)(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

    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

    题解:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #include <set>
    #include <map>
    #define LL long long
    #define INF 0x3f3f3f3f
    using namespace std;
    const int maxn=505;
    int n,m,c1,c2;
    int vis[maxn],num[maxn];
    int e[maxn][maxn];
    int dis[maxn];
    int w[maxn],weight[maxn];
    void init(){
        memset(vis,0,sizeof vis);
        memset(dis,INF,sizeof dis);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i==j) e[i][j]=0;
                else e[i][j]=INF;
            }
        }
    }
    void Dijkstra(){
        for(int i=0;i<n;i++){
            dis[i]=e[c1][i];
        }
        dis[c1]=0;
        //vis[c1]=1;
        num[c1]=1;
        w[c1]=weight[c1];
        for(int i=0;i<n;i++){
            int minx=INF;
            int index=-1;
            for(int j=0;j<n;j++){
                if(!vis[j]&&dis[j]<minx){
                    minx=dis[j];
                    index=j;
                }
            }
            if(index==-1) return;
            vis[index]=1;
            for(int j=0;j<n;j++){
                if(!vis[j]&&e[index][j]<INF){
                    if(dis[index]+e[index][j]<dis[j]){
                        dis[j]=dis[index]+e[index][j];
                        num[j]=num[index];
                        w[j]=w[index]+weight[j];
                    }else if(dis[index]+e[index][j]==dis[j]){
                        num[j]=num[j]+num[index];
                        if(w[index]+weight[j]>w[j]){
                            w[j]=w[index]+weight[j];
                        }
                    }
                }
            }
        }
    
    }
    int main()
    {
        scanf("%d%d%d%d",&n,&m,&c1,&c2);
        init();
        for(int i=0;i<n;i++){
            scanf("%d",&weight[i]);
        }
        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;
        }
        Dijkstra();
        printf("%d %d
    ",num[c2],w[c2]);
        return 0;
    }
    


  • 相关阅读:
    SQL分页语句
    读、写 节点 XML方法总结
    新手入门:XHTML DHTML SHTML的区别
    ROS映射非21端口的FTP服务器设置
    图解:asp.net三种重定向方法
    在ASP.NET中显示进度条
    C#读取文本文件
    Net3.5都快来了,.Net2.0你们都知道多少呢?
    div+css布局漫谈
    自定义的向客户端输出Javascript脚本alert函数
  • 原文地址:https://www.cnblogs.com/kzbin/p/9205210.html
Copyright © 2011-2022 走看看