Luogu P1613 Running Road Problem Solution

topic

This problem has a small n, so floyd can also be competent for the shortest time;

Let g[i][j][k] represent that there is a path with weight k from i to j at present;

Since the edge weights are all 1, g[i][j][k]=g[i][p][k-1]+g[p][j][k-1];

For f[i][j]; if i to j can be passed in 1 step, then = 1; otherwise = inf;

Then just run floyd;

#include <bits/stdc++.h>
using namespace std;
int b[110][110][40];
int f[110][110];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(register int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		b[u][v][0]=1;
		f[u][v]=1;
	}
	for(int k=1;k<=30;k++){
		for(int p=1;p<=n;p++){
			for(int i=1;i<=n;i++){
				for(int j=1;j<=n;j++){
					if(b[i][p][k-1]&&b[p][j][k-1]){
						b[i][j][k]=1;
						f[i][j]=1;
					}
				}
			}
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(f[i][j]!=1) f[i][j]=999999999;
		}
	}
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
			}
		}
	}
	cout<<f[1][n];
}

 

Reprinted in: https://www.cnblogs.com/kamimxr/p/11607751.html