Einbahnstrasse HDU - 2923 (Floyd， 注意建图)

题目描述

Einbahnstra  e (German for a one-way street) is a street on which vehicles should only move in one direction. One reason for having one-way streets is to facilitate a smoother flow of traffic through crowded areas. This is useful in city centers, especially old cities like Cairo and Damascus. Careful planning guarantees that you can get to any location starting from any point. Nevertheless, drivers must carefully plan their route in order to avoid prolonging their trip due to one-way streets. Experienced drivers know that there are multiple paths to travel between any two locations. Not only that, there might be multiple roads between the same two locations. Knowing the shortest way between any two locations is a must! This is even more important when driving vehicles that are hard to maneuver (garbage trucks, towing trucks, etc.) 

You just started a new job at a car-towing company. The company has a number of towing trucks parked at the company's garage. A tow-truck lifts the front or back wheels of a broken car in order to pull it straight back to the company's garage. You receive calls from various parts of the city about broken cars that need to be towed. The cars have to be towed in the same order as you receive the calls. Your job is to advise the tow-truck drivers regarding the shortest way in order to collect all broken cars back in to the company's garage. At the end of the day, you have to report to the management the total distance traveled by the trucks.

题目大意， 输入 A城市  （方向， 费用） B城市 ， 输出最短路。 A <-X  B 表示可以从B城市到A城市 长度为X. 其他同理

思路： 先用map 讲所有城市转换成数字的形式来表示， 输入中间用string 来存， 然后读取方向， 读取长度。 最后跑一下floyd

AC code:

 #include <iostream>
 #include <cstring>
 #include <string>
 #include <map>
 using namespace std;
 #define INF 0x3f3f3f3f
 int mp[1100][1100], cnt;
 map<string, int> city;
 
 int main(void) {
    ios::sync_with_stdio(false);
    int n, c, m, time = 0;
    while (cin >> n >> c >> m && n+m+c) {
        city.clear();
        memset(mp, INF, sizeof(mp));
        string s[1100];
        cnt = 0;
        for (int i = 0; i <= c; i++) {
            cin >> s[i];
            if (!city[s[i]]) city[s[i]] = ++cnt;
        }
        string s1, s2, s3;
        for (int i = 0; i < m; i++) {
            cin >> s1 >> s2 >> s3;
            if (!city[s1]) city[s1] = ++cnt;
            if (!city[s3]) city[s3] = ++cnt;  // 将城市（字符串）转化成 编号（数字）
            
            int num = 0;
            for (int j = 0; j < s2.size(); j++) {
                if (s2[j] >= '0' && s2[j] <= '9') {
                    num = num*10 + (s2[j] - '0');  // 读取中间字符串的数字
                }
            }
            
            if (s2[0] == '<')  // 建图
                if (num < mp[city[s3]][city[s1]]) 
                    mp[city[s3]][city[s1]] = num;
            if (s2[s2.size()-1] == '>') 
                if (num < mp[city[s1]][city[s3]]) 
                    mp[city[s1]][city[s3]] = num;
        }
           
        for (int k = 1; k <= n; k++) {  //floyd
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    if (mp[i][j] > mp[i][k] + mp[k][j])
                        mp[i][j] = mp[i][k] + mp[k][j];  
                }
            }
        }
        
        int ans = 0;
        for (int i = 1; i <= c; i++) {
            ans += mp[1][city[s[i]]] + mp[city[s[i]]][1];
        }
        cout << ++time << ". " << ans << endl;
    }
    return 0;
 }