Sicily 1031. Campus

1031. Campus

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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

Problem Source

ZSUACM Team Member

Dijkstra，没什么好说的

这题神坑的地方就是：如果两个地点一个在列表中一个不在列表中，输出-1

两个都不在列表中但是两个地点相同输出0

注意判定。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
#define rep(i,l,r) for(int i = l; i <= r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for(Edge *p = last[x]; p; p = p -> pre)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 210;
inline int read(){
    int ans = 0, f = 1; char c = getchar();
    for(; !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for(; isdigit(c); c = getchar()) ans = ans * 10 + c - '0';
    return ans * f;
}
struct Edge{
    Edge *pre; int to, cost;
}edge[510];
Edge *pt, *last[maxn];
struct Node{
    int x, d;
    Node(int _x,int _d) : x(_x), d(_d){    }
    inline bool operator < (const Node &_Tp) const{
        return d > _Tp.d;
    }
};
priority_queue <Node> q;
string si, ti, s, t;
int n, cnt, S, T, d[maxn];
inline void addedge(int x,int y,int z){
    pt->pre = last[x]; pt->to = y; pt->cost = z; last[x] = pt++; 
}
void dijkstra(){
    while (!q.empty()) q.pop();
    clr(d,INF); d[S] = 0;
    q.push(Node(S,0));
    while (!q.empty()){
        Node now = q.top(); q.pop();
        if (d[now.x] != now.d) continue;
        travel(now.x){
            if (d[p->to] > d[now.x] + p->cost){
                d[p->to] = d[now.x] + p->cost;
                q.push(Node(p->to,d[p->to])); 
            }
        }
    }
}
void work(){
    clr(last,0); pt = edge; cnt = 0;
    n = read();
    map <string, int> list;
    rep(i,1,n){
        cin >> si >> ti;
        int x = read();
        if (!list.count(si)) list.insert(pair <string,int> (si,++cnt));
        if (!list.count(ti)) list.insert(pair <string,int> (ti,++cnt));
        addedge(list[si],list[ti],x);
        addedge(list[ti],list[si],x);
    }
    cin >> s >> t;
    if ((!list.count(s)) || (!list.count(t))){
        if (s != t){
            printf("-1\n"); return;
        }
        else printf("0\n"); return;
    }
    S = list[s]; T = list[t];
    if (S == T){
        printf("0\n"); return;
    }
    dijkstra();
    if (d[T] == INF) printf("-1\n"); else printf("%d\n",d[T]);
}
int main(){
    int C = read();
    while (C--) work();
    return 0;
}