Time limit: 1 s

Space constraints: 128000 KB

Topic Level: Master

Description

H City is a tourist attraction, and thousands of people come to visit it every year. To facilitate tourists, bus companies have set up bus stops and opened some one-way bus routes in various tourist attractions, hotels and other places. Each one-way bus route starts from a bus stop and passes through several bus stops in turn, finally arriving at the terminal bus stop.

A traveler recently visited H City. He would like to visit S Park. But if there is no bus from his hotel to reach S Park directly, he may have to take one bus for several stops and then come down to another bus on the same platform, so he can get to S Park several times later.

Now use integers 1, 2, and ____________. N. Number all bus stops in H City and agree that the passenger's hotel bus stop number is 1. The number of S Park Bus Station is N.

Write a program to help the passenger find an optimal ride plan, so that he can change at least the number of times during the ride from the hotel to S Park.

Input Description

The input file is INPUT.TXT. The first line of the document has two digits M and N (1<=M<=100.1<N<=500), indicating that M one-way bus routes have been opened, with a total of N stations. Information on bus routes from line 1 to M is given in turn from line 2 to issue M. Line i+1 gives the information of the bus route i. From left to right, it gives the information of all station numbers on the bus line which are separated by a space between two adjacent station numbers.

Output Description

The output file is OUTPUT.TXT, with only one line. If you can't take a bus from the hotel to S Park, then output "N0", otherwise output the minimum number of changes found by your program. The number of changes is 0, which means that you can arrive without changing trains.

Sample Input

3 7

6 7

4 7 3 6

2 1 3 5

Sample Output

2

Data Size & Hint

 

Thoughts: Read in data + Floyed violence using read-in optimization

Be careful of the details.

Benni debugged for three hours because of two i and one n

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[510][510];
int map[510][510];
int cd[510];
int inf=0x7fffffff;
int main()
{
    
    for(int i=0;i<=501;i++)
    for(int j=0;j<=501;j++)
    map[i][j]=inf;
    //memset(map,0x7f,sizeof(map));
    //for(int i=1;i<=)
    int n,m;
    //scanf("%d%d",&m,&n);
    cin>>m>>n;
    char p;
    for(int i=1;i<=m;i++)
    {
        int now=1;
        char p;
        while(1)
        {
            int bc=0;//Storage results
            p=getchar();
            while(p>='0'&&p<='9')
            {
                bc=bc*10+p-'0';
                p=getchar();
            }
            if(bc!=0)
            {
                a[i][now]=bc;
                now++;
            }
            
            if(p==' ')
            continue;
            if(p=='\n')
            {
                if(a[i][now-1]==0)
                continue;
                cd[i]=now;
                now=1;
                break;
            }
        }
        cd[i]--;
    }
    for(int k=1;k<=m;k++)
    {
        for(int i=1;i<=cd[k];i++)
        {
            for(int j=i+1;j<=cd[k];j++)
            {
                if(i==j)
                map[a[k][i]][a[k][j]]=0;
                else
                map[a[k][i]][a[k][j]]=1;
            }
        }
    }
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(map[i][k]!=inf&&map[k][j]!=inf&&((map[i][k]+map[k][j]!=0)))
                if(map[i][j]>map[i][k]+map[k][j])
                {
                    map[i][j]=map[i][k]+map[k][j];
                }
            }
        }
    }
    if(map[1][n]==inf)
    {
        cout<<"NO";
    }
    else
    {
        cout<<map[1][n]-1;
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[510][510];
int map[510][510];
int cd[510];
int inf=0x7fffffff;
int main()
{
    
    for(int i=0;i<=501;i++)
    for(int j=0;j<=501;j++)
    map[i][j]=inf;
    //memset(map,0x7f,sizeof(map));
    //for(int i=1;i<=)
    int n,m;
    //scanf("%d%d",&m,&n);
    cin>>m>>n;
    char p;
    for(int i=1;i<=m;i++)
    {
        int now=1;
        char p;
        while(1)
        {
            int bc=0;//Storage results
            p=getchar();
            while(p>='0'&&p<='9')
            {
                bc=bc*10+p-'0';
                p=getchar();
            }
            if(bc!=0)
            {
                a[i][now]=bc;
                now++;
            }
            
            if(p==' ')
            continue;
            if(p=='\n')
            {
                if(a[i][now-1]==0)
                continue;
                cd[i]=now;
                now=1;
                break;
            }
        }
        cd[i]--;
    }
    for(int k=1;k<=m;k++)
    {
        for(int i=1;i<=cd[k];i++)
        {
            for(int j=i+1;j<=cd[k];j++)
            {
                if(i==j)
                map[a[k][i]][a[k][j]]=0;
                else
                map[a[k][i]][a[k][j]]=1;
            }
        }
    }
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(map[i][k]!=inf&&map[k][j]!=inf&&((map[i][k]+map[k][j]!=0)))
                if(map[i][j]>map[i][k]+map[k][j])
                {
                    map[i][j]=map[i][k]+map[k][j];
                }
            }
        }
    }
    if(map[1][n]==inf)
    {
        cout<<"NO";
    }
    else
    {
        cout<<map[1][n]-1;
    }
    return 0;
}