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+10;
const int MOD = 1e9+7;
#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=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;}
const int M =35;

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=0;i<n;i++) scanf("%d",&f[i][0]);
    for(int i=0;i<n;i++) {scanf("%d",&sum[i][0]);mn[i][0]=sum[i][0];}
    for(int j=1;j<M;j++){
        for(int i=0;i<n;i++){
            f[i][j]=f[f[i][j-1]][j-1];
            sum[i][j]=sum[f[i][j-1]][j-1]+sum[i][j-1];
            mn[i][j]=min(mn[i][j-1],mn[f[i][j-1]][j-1]);
        }
    }
    LL ans;
    for(int i=0;i<n;i++){
        int v=i,minn=inf;
        LL K=k;
        ans=0;
        for(int j=M-1;j>=0;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
",ans,minn);
    }
    return 0;
}