sicily 1031 Campus（图算法）

Posted 2020-08-27 zmj97

Description

       At present, Zhongshan University has 4 campuses with a total area of 6.17 square kilometers sitting respectively on both sides of the Pearl River or facing the South China Sea. The Guangzhou South Campus covers an area of 1.17 square kilometers, the North Campus covers an area of 0.39 square kilometers, the Guangzhou East Campus has an area of 1.13 square kilometers and the Zhuhai Campus covers an area of 3.48 square kilometers. All campuses have exuberance of green trees, abundance of lawns and beautiful sceneries, and are ideal for molding the temperaments, studying and doing research.

 

       Sometime, the professors and students have to go from one place to another place in one campus or between campuses. They want to find the shortest path between their source place S and target place T. Can you help them?

 

Input

The first line of the input is a positive integer C. C is the number of test cases followed. In each test case, the first line is a positive integer N (0<N<=100) that represents the number of roads. After that, N lines follow. The i-th(1<=i<=N) line contains two strings Si, Ti and one integer Di (0<=Di<=100). It means that there is a road whose length is Di between Si and Ti. Finally, there are two strings S and T, you have to find the shortest path between S and T. S, T, Si(1<=i<=N) and Ti(1<=i<=N) are all given in the following format: str_Campus.str_Place. str_Campus represents the name of the campus, and str_Place represents the place in str_Campus. str_Campus is "North", "South", "East" or "Zhuhai". str_Place is a string which has less than one hundred lowercase characters from "a-z". You can assume that there is at most one road directly between any two places.

Output

The output of the program should consist of C lines, one line for each test case. For each test case, the output is a single line containing one integer. If there is a path between S and T, output the length of the shortest path between them. Otherwise just output "-1" (without quotation mark). No redundant spaces are needed.

Sample Input

1
2
South.xiaolitang South.xiongdelong 2
South.xiongdelong Zhuhai.liyuan 100
South.xiongdelong South.xiaolitang

Sample Output

2

 

使用dijkstra算法，算法思路可以看https://www.youtube.com/watch?v=gdmfOwyQlcI

因为dijkstra第三个参数传错debug了好久，以后要注意细节。

看到别人用了map来计算新城市，我是直接暴力查找添加的。

 

 

以下是代码：

#include <iostream>
#include <string>
using namespace std;

#define INF 1000000
#define MAX 210
int roadLength[MAX][MAX];
string cities[MAX];
bool visited[MAX];
int len[MAX];

void initial(int n) {        // initial all arrays
    for (int i = 1; i <= n; i++) {
        cities[i] = "";
        for (int j = 1; j <= n; j++) roadLength[i][j] = INF;
        roadLength[i][i] = 0;
        visited[i] = false;
        len[i] = INF;
    }
}

int cityPos(string x, int cityCount) {        // return city pos in array cities, if not exist, return -1
    for (int i = 1; i <= cityCount; i++) {
        if (cities[i] == x) return i;
    }
    return -1;
}

void addCity(string x, int &xpos, int &cityCount) {        // if the city not exist in the array, add it;  xpos store the pos of the city
    xpos = cityPos(x, cityCount);
    if (xpos == -1) {
        cities[++cityCount] = x;
        xpos = cityCount;    
    }    
}

int dijkstra(int startCityPos, int endCityPos, int n) {    // n is cityCount
    len[startCityPos] = 0;
    for (int i = 1; i <= n; i++) {
        // currentVisitPos is the pos of the city which is not visited and has the shortest len
        int currentVisitPos = startCityPos;
        int minLen = INF;
        for (int j = 1; j <= n; j++) {
            if (!visited[j] && len[j] < minLen) {
                minLen = len[j];
                currentVisitPos = j;
            }
        }
        visited[currentVisitPos] = true;
        // update the lens of unvisited cities
        for (int j = 1; j <= n; j++) {
            if (!visited[j] && len[currentVisitPos] + roadLength[currentVisitPos][j] < len[j]) {
                len[j] = len[currentVisitPos] + roadLength[currentVisitPos][j];
            }
        }
    }
    if (visited[endCityPos]) return len[endCityPos];
    return -1;
}

int main() {
    int t;
    cin>>t;
    while(t--) {
        int n;
        cin>>n;
        initial(n*2);
        int cityCount = 0;
        for (int i = 0; i < n; i++) {
            string x, y;
            int length, xpos, ypos;
            cin>>x>>y>>length;
            addCity(x, xpos, cityCount);
            addCity(y, ypos, cityCount);
            roadLength[xpos][ypos] = roadLength[ypos][xpos] = length;
        }
        string startCity, endCity;
        cin>>startCity>>endCity;
        int startCityPos = cityPos(startCity, cityCount), endCityPos = cityPos(endCity, cityCount);
        if (startCity == endCity) cout<<0<<endl;
        else if (startCityPos == -1 || endCityPos == -1) cout<<-1<<endl;
        else cout<<dijkstra(startCityPos, endCityPos, cityCount)<<endl;
    }
    return 0;
}