poj-1847-Tram(最短路问题)

传送门

Description

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”.

Sample Input

3 2 1
2 2 3
2 3 1
2 1 2

Sample Output

0

---

题意：
就是有n个交叉点，就当做有n个点就行，然后这些点和其他点有些路径，每个点是一个开关，开关只能有一个方向走一条路，而第一个数就是默认的开关指向，不用旋转。

默认的指向实际上只需要旋转0次，而其他路径只需要旋转1次，无论是哪条，只需1次

---

代码：

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>const long long N = 100+10, MAX = 200000000;using namespace std;int n, a, b, map[N][N], dis[N], vis[N];void Dijkstra(){    int i, j, u, m;    for (i = 1; i <= n; i++)    {        vis[i] = 0;        dis[i] = map[a][i];    }    vis[a] = 1;    dis[a] = 0;    for (i = 1; i < n; i++)    {        u = -1;        m = MAX;        for (j = 1; j <= n; j++)        {            if (!vis[j] && m > dis[j])            {                m = dis[j];                u = j;            }        }        vis[u] = 1;        for (j = 1; j <= n; j++)        {            if (!vis[j] && dis[j] > dis[u]+map[u][j])            {                dis[j] = map[u][j] + dis[u];            }        }    }}int main(){#ifndef ONLINE_JUDGE    freopen("1.txt", "r", stdin);#endif    int i, j, k, t, x;    while(cin >> n >> a >> b)    {        for (i = 0; i <= n; i++)        {            for (j = 0; j <= n; j++)            {                map[i][j] = MAX;            }        }        for (i = 1; i <= n; i++)        {            cin >> t;            for (j = 0; j < t; j++)            {                cin >> x;                if (j)  map[i][x] = 1;                else map[i][x] = 0;            }        }        Dijkstra();        if (dis[b] != MAX)            cout << dis[b] << endl;        else            cout << "-1" << endl;    }    return 0;}