1031. Campus

1031. Campus

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

12South.xiaolitang South.xiongdelong 2South.xiongdelong Zhuhai.liyuan 100South.xiongdelong South.xiaolitang

Sample Output

2

Problem Source

ZSUACM Team Member

#include <iostream>#include <map>#include <string>#include <cstring>using namespace std;const int INF=100000;//自訂無窮大int adj[205][205];//記錄兩點間距離bool flag[205];//標記處理過的點int dis[205];//記錄各點到某點的距離int t,road,d,n;//t為測試資料群組數,road為每組資料行數,d為沒行資料表示的兩點間距離,n記錄映射對應的路的編號string start,send;//每行輸入的兩條路int dijkstra(int a,int b)//dijkstra法處理a到b的最短路{memset(flag,0,sizeof(flag));for(int i=0;i<n;i++)dis[i]=((i==a)?0:INF);//其他各點初始化距起點為INFfor(int i=0;i<n;i++){int min=INF,x=a;///x為要求出的未標記點中距離a最近的點for(int y=0;y<n;y++){if(!flag[y]&&dis[y]<min){min=dis[y];x=y;}}flag[x]=1;//最近的點標記，下次則從未標記的點中選取for(int y=0;y<n;y++)dis[y]=(dis[y]>dis[x]+adj[x][y]?dis[x]+adj[x][y]:dis[y]);}if(flag[b])//若a點與b點連通return dis[b];else return -1;}int main(){cin>>t;while(t--){cin>>road;for(int i=0;i<205;i++)for(int j=0;j<205;j++)adj[i][j]=((i==j)?0:INF);map<string,int>mymap;//路與編號相對應n=0;//n記錄對應路的編號for(int i=0;i<road;i++){//string end;cin>>start>>send>>d;if(!mymap.count(start))mymap[start]=n++;if(!mymap.count(send))mymap[send]=n++;adj[mymap[start]][mymap[send]]=adj[mymap[send]][mymap[start]]=d;}cin>>start>>send;if(start==send) cout<<0<<endl;else if(!mymap.count(start)||!mymap.count(send)) cout<<-1<<endl;else cout<<dijkstra(mymap[start],mymap[send])<<endl;}return 0;}