P1613 foot problem solution

Today, there's really nothing~~Then I'll sell Meng bar (~ ¯ ▽ ¯) ~~~

Luogu

A thinking problem~~It may be only for me, for the bigwigs are evident~~

Simplify the meaning of problems

Give you a map, find a path to make it bit paths \ (1 \) minimum number of

analysis

This topic question, there are two pit~~Perhaps only for me~~

Every time to go, only you can stop at the endpoint, that is, if there is a road from \ (i \) to \ (J \) , if its path length of 7, not from \ (i \) go directly to \ (J \) , you go \ (3 \) times.
Since \ (1 \) so directly run the shortest path is not necessarily the best.

Then think of solution

First, since the label ~ / path is too long special / cow * ~~ machine of the way on foot, so we thought of using multiplication to solve this problem

Then consider this last question is the most short-circuit requirements. We can see the data range \ (the n-<= 50 \) , so I think Freud, find the shortest path. However, due to the above conclusions, we can not directly run the shortest, and therefore we need to do a conversion on this map. This question is difficult to make the conversion to the point figure, so we chose to make the conversion to edge / path.

Then consider the transformation of the path. The reason can not directly run the Shortest Path before considering: We can not determine how to direct the path of binary bits in a relatively short period of time \ (1 \) the number of less, so when we put it into the path of contribution in binary in \ (1 \) the shortest number you can count.

Therefore, first pre-processing from \ (I \) to \ (J \) shortest path (1 \) \ number, and then seek the shortest side of the like.

For seeking \ (I \) to \ (J \) shortest path \ (1 \) number, since \ (I \) to \ (J \) path has a property~~Oh, I do not know the specific name~~: \ (I \) to \ (T \) in the number of \ (K \) , \ (T \) to \ (J \) in the number of \ (K \) , then the \ (I \) to \ (J \) of \ (1 \) number is \ (K + 1 \) . Therefore, we can use similar transmitted shortest way between each two points to find closure in bins \ (1 \) number

code

#include<iostream>
#include<cstring>
using namespace std;
int n,m,x,y,dis[55][55];
bool map[55][55][65];
inline void pre(){
    for(register int t=0;t<=64;t++)
        for(register int k=1;k<=n;k++)
            for(register int i=1;i<=n;i++)
                for(register int j=1;j<=n;j++)
                    if(map[i][k][t-1]&&map[k][j][t-1])
                        map[i][j][t]=1,dis[i][j]=1;
}
inline void floyed(){
    for(register int k=1;k<=n;k++)
        for(register int i=1;i<=n;i++)
            for(register int j=1;j<=n;j++)
                dis[i][j]=min(dis[i][k]+dis[k][j],dis[i][j]);
}
int main(){
    scanf("%d%d",&n,&m);
    memset(dis,0x3f,sizeof(dis));
    for(register int i=1;i<=m;i++)
        scanf("%d%d",&x,&y),
        dis[x][y]=1,map[x][y][0]=1;
    pre();floyed();printf("%d",dis[1][n]);
}

Finally, international practice, thankyou for your attention