codeforce 702E Analysis of Pathes in Functional Graph RMQ+二进制

http://codeforces.com/contest/702

题意：n个点，n条边，每个点出边只有一条，问从每个点出发经过k条边的边权和，以及边权最小值

思路：

f[i][j] 第i个点出发，经过2^j条边后的相连点 其余类似

二进制巧妙解决问题应用太广了

 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <sstream>
 #include <string>
 #include <algorithm>
 #include <list>
 #include <map>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <cstdlib>
 // #include <conio.h>
 using namespace std;
 #define pi acos(-1.0)
 const int N = 1e5+;
 const int MOD = 1e9+;
 #define inf 0x7fffffff
 typedef long long  LL;

 void frein(){freopen("in.txt","r",stdin);}
 void freout(){freopen("out.txt","w",stdout);}
 inline LL read(){LL x=,f=;char ch=getchar();while(ch>''||ch<'') {if(ch=='-') f=-;ch=getchar();}while(ch>=''&&ch<='') { x=x*+ch-'';ch=getchar();}return x*f;}
 const int M =;

 int f[N][M];
 LL sum[N][M];
 int mn[N][M];
 int main(){
     int n;
     LL k;
     n=read(),k=read();
     for(int i=;i<n;i++) scanf("%d",&f[i][]);
     for(int i=;i<n;i++) {scanf("%d",&sum[i][]);mn[i][]=sum[i][];}
     for(int j=;j<M;j++){
         for(int i=;i<n;i++){
             f[i][j]=f[f[i][j-]][j-];
             sum[i][j]=sum[f[i][j-]][j-]+sum[i][j-];
             mn[i][j]=min(mn[i][j-],mn[f[i][j-]][j-]);
         }
     }
     LL ans;
     for(int i=;i<n;i++){
         int v=i,minn=inf;
         LL K=k;
         ans=;
         for(int j=M-;j>=;j--){
             if((1LL<<j)<=K){
                 ans+=sum[v][j];
                 minn=min(minn,mn[v][j]);
                 v=f[v][j];
                 K-=(1LL<<j);
             }
         }
         printf("%I64d %d\n",ans,minn);
     }
     return ;
 }