POJ 1847 Tram （最短路）

Tram

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/N

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

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

Hint

##题意：

求图中#A到#B的最短路.


##题解：

裸的最短路，数据非常小，随便求就好.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 110
#define mod 1000000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,s,t;

int dis[maxn][maxn];

void init() {

for(int i=1; i<=n; i++)

for(int j=1; j<=n; j++)

dis[i][j] = (i==j? 0:inf);

}

void floyd() {

for(int k=1; k<=n; k++)

for(int i=1; i<=n; i++)

for(int j=1; j<=n; j++)

if(dis[i][j] > dis[i][k]+dis[k][j])

dis[i][j] = dis[i][k] + dis[k][j];

}

int main(void)

{

//IN;

while(scanf("%d %d %d", &n,&s,&t) != EOF)
{
    init();
    for(int i=1; i<=n; i++) {
        int m; scanf("%d", &m);
        for(int j=1; j<=m; j++) {
            int x; scanf("%d", &x);
            dis[i][x] = j==1? 0:1;
        }
    }

    floyd();

    if(dis[s][t] == inf) printf("-1\n");
    else printf("%d\n", dis[s][t]);
}

return 0;

}