POJ1847 Tram

　　　　　　　　　　　　　　　　　 Tram

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20274		Accepted: 7553

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

Dijkstra+堆优化，比较简单的一道题。

说一下思路吧：
每个节点与所有可到达节点之间连边，与初始指向节点的权值为0，与其余可到达的节点的权值为1。
然后求最短路。

#include <cstdio>
#include <queue>
using namespace std;

inline int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0' || c>'9'){if (c=='-')f=-1;c=getchar();}
	while (c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-48;c=getchar();}
	return x*f;
}

inline void print(int x)
{
	if (x<0)x=-x,putchar('-');
	if (x>9)print(x/10);
	putchar(x%10+48);
}

inline void print(int x,char c)
{
	print(x);
	putchar(c);
}

const int INF=10000000;
const int MAXN=101;
int n,from,to;
int a[MAXN];
int cost[MAXN][MAXN];

struct dij
{
	int id,dis;
	bool operator < (const dij tmp) const
	{
		return dis>tmp.dis;
	}
};
int dis[MAXN];

inline void dijkstra()
{
	int u;
	for (int i=1;i<=n;i++)dis[i]=INF;
	dis[from]=0;
	priority_queue<dij> Q;
	Q.push((dij){from,0});
	while (!Q.empty())
	{
		u=Q.top().id;Q.pop();
		for (int i=1;i<=n;i++)
			if (dis[u]+cost[u][i]<dis[i])
			{
				dis[i]=dis[u]+cost[u][i];
				Q.push((dij){i,dis[i]});
			}
	}
}

int main()
{
	n=read();from=read();to=read();
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			cost[i][j]=INF;
	for (int i=1;i<=n;i++)cost[i][i]=0;
	int top,k;
	for (int i=1;i<=n;i++)
	{
		top=read();
		if (!top)continue;
		cost[i][read()]=0;
		for (int j=2;j<=top;j++)cost[i][read()]=1;
	}
	dijkstra();
	if (dis[to]<INF)print(dis[to],'\n');
	else print(-1,'\n');
	return 0;
}

 

转载于:https://www.cnblogs.com/lzxzy-blog/p/10501833.html