【题解】Tram POJ - 1847 ⭐⭐⭐ 【简单建图】

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在Zagreb的电车网络中，每个交叉口都连接着多条轨道，且默认指向第一条轨道。任务是找到从交叉口A到B的路径，以最小化手动切换轨道次数。通过构建图形并使用Dijkstra算法，可以确定从A到B的最短路径。

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Tram POJ - 1847

Tram network in Zagreb consists of a number of intersections and rails connecting some of them. In every intersection there is a switch pointing to the one of the rails going out of the intersection. When the tram enters the intersection it can leave only in the direction the switch is pointing. If the driver wants to go some other way, he/she has to manually change the switch.

When a driver has do drive from intersection A to the intersection B he/she tries to choose the route that will minimize the number of times he/she will have to change the switches manually.

Write a program that will calculate the minimal number of switch changes necessary to travel from intersection A to intersection B.

Input

The first line of the input contains integers N, A and B, separated by a single blank character, 2 <= N <= 100, 1 <= A, B <= N, N is the number of intersections in the network, and intersections are numbered from 1 to N.

Each of the following N lines contain a sequence of integers separated by a single blank character. First number in the i-th line, Ki (0 <= Ki <= N-1), represents the number of rails going out of the i-th intersection. Next Ki numbers represents the intersections directly connected to the i-th intersection.Switch in the i-th intersection is initially pointing in the direction of the first intersection listed.

Output

The first and only line of the output should contain the target minimal number. If there is no route from A to B the line should contain the integer “-1”.

Examples

Sample Input
3 2 1
2 2 3
2 3 1
2 1 2
Sample Output
0

Hint

题意:

在一个交通网络上有N个路口, 每个路口指向多个方向, 默认驶向第一个方向, 驶向其他方向时需要进行一次操作, 求从a到b最小的操作数

题解:

直接建图, 默认权值为inf, 有操作为1, 默认为0

经验小结:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define ms(x, n) memset(x,n,sizeof(x));
typedef  long long LL;
const int inf = 1<<30;
const LL maxn = 110;

int N, s, e, w[maxn][maxn];

int d[maxn];
bool used[maxn];
typedef pair<int, int> P;
void Dijkstra(){
    priority_queue<P, vector<P>, greater<P> > q;
    fill(d, d+maxn, inf);
    q.push(P(d[s]=0, s));
    while(!q.empty()){
        P cur = q.top();
        q.pop();
        int u = cur.second;
        if(used[u]) continue;
        used[u] = true;
        for(int v = 1; v <= N; ++v){
            if(d[u]+w[u][v] < d[v]){
                d[v] = d[u]+w[u][v];
                q.push(P(d[v], v));
            }
        }
    }
}
int main()
{
    int k, v;
    cin >> N >> s >> e;
    fill(w[0], w[0]+maxn*maxn, inf);
    for(int u = 1; u <= N; ++u){
        cin >> k;
        for(int j = 1; j <= k; ++j){
            cin >> v;
            if(j==1) w[u][v] = 0;
            else w[u][v] = 1;
        }
    }
    Dijkstra();
    if(d[e]<inf) cout << d[e] << endl;
    else cout << "-1" << endl;

	return 0;
}