Luogu P1613 foot solution to a problem report

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[Title] effect

[Analysis] ideas

We set $ g [i] [j] [k] $ represents from $ I $ take $ 2 ^ k $ steps can reach $ j $, $ d [i] [j] $ represents $ I $ to $ J $ happened how many seconds to go.

Pretreatment with a multiplication $ g $, $ can then run the Floyd $ shortest it! QwQ

【Code】

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #define g() getchar()
 8 #define rg register
 9 #define go(i,a,b) for(rg int i=a;i<=b;i++)
10 #define back(i,a,b) for(rg int i=a;i>=b;i--)
11 #define db double
12 #define ll long long
13 #define il inline
14 #define pf printf
15 #define mem(a,b) memset(a,b,sizeof(a))
16 using namespace std;
17 int fr(){
18     int w=0,q=1;
19     char ch=g();
20     while(ch<'0'||ch>'9'){
21         if(ch=='-') q=-1;
22         ch=g();
23     }
24     while(ch>='0'&&ch<='9') w=(w<<1)+(w<<3)+ch-'0',ch=g();
25     return w*q;
26 }
27 const int N=52;
28 int n,m,d[N][N];
29 bool f[N][N][65];
30 int main(){
31     //freopen("","r",stdin);
32     //freopen("","w",stdout);
33     n=fr();m=fr();mem(d,0x3f);
34     go(i,1,m){
35         rg int x=fr(),y=fr();
36         d[x][y]=1;f[x][y][0]=1;
37     }
38     go(k,1,64) go(i,1,n) go(t,1,n) go(j,1,n)//注意循环嵌套的顺序
39         if(f[i][t][k-1]&&f[t][j][k-1]) f[i][j][k]=1,d[i][j]=1;
40     go(k,1,n) go(i,1,n) go(j,1,n) d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
41     pf("%d\n",d[1][n]);    
42     return 0;
43 }

Code poke here