codeforces 702E Analysis of Pathes in Functional Graph(倍增)

E. Analysis of Pathes in Functional Graph

 

You are given a functional graph. It is a directed graph, in which from each vertex goes exactly one arc. The vertices are numerated from 0 to n - 1.

Graph is given as the array f0, f1, ..., fn - 1, where fi — the number of vertex to which goes the only arc from the vertex i. Besides you are given array with weights of the arcs w0, w1, ..., wn - 1, where wi — the arc weight from i to fi.

The graph from the first sample test.

Also you are given the integer k (the length of the path) and you need to find for each vertex two numbers si and mi, where:

  • si — the sum of the weights of all arcs of the path with length equals to k which starts from the vertex i;
  • mi — the minimal weight from all arcs on the path with length k which starts from the vertex i.

The length of the path is the number of arcs on this path.

Input

The first line contains two integers n, k (1 ≤ n ≤ 105, 1 ≤ k ≤ 1010). The second line contains the sequence f0, f1, ..., fn - 1 (0 ≤ fi < n) and the third — the sequence w0, w1, ..., wn - 1 (0 ≤ wi ≤ 108).

Output

Print n lines, the pair of integers simi in each line.

Examples
input
7 3
1 2 3 4 3 2 6
6 3 1 4 2 2 3
output
10 1
8 1
7 1
10 2
8 2
7 1
9 3
input
4 4
0 1 2 3
0 1 2 3
output
0 0
4 1
8 2
12 3
input
5 3
1 2 3 4 0
4 1 2 14 3
output
7 1
17 1
19 2
21 3
8 1

#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<LL,int>pii;
const int N=1e5+5;
LL ws[N][40],k;
int f[N][40],wm[N][40],n;
void bz()
{
    for(int j=1;(1LL<<j)<=k;j++)
        for(int i=0;i<n;i++)
        {
            f[i][j]=f[f[i][j-1]][j-1];
            ws[i][j]=ws[i][j-1]+ws[f[i][j-1]][j-1];
            wm[i][j]=min(wm[i][j-1],wm[f[i][j-1]][j-1]);
        }
}
pii query(int u)
{
    pii res;
    res.fi=0,res.se=1e8;
    for(int i=0;i<40;i++)
        if(k&(1LL<<i))
            res.fi+=ws[u][i],res.se=min(wm[u][i],res.se),u=f[u][i];
    return res;
}
int main()
{
    scanf("%d%I64d",&n,&k);
    for(int i=0;i<n;i++)
        scanf("%d",&f[i][0]);
    for(int i=0;i<n;i++)
        scanf("%d",&wm[i][0]),ws[i][0]=wm[i][0];
    bz();
    for(int i=0;i<n;i++)
    {
        pii ans=query(i);
        printf("%I64d %d
",ans.fi,ans.se);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/homura/p/5720837.html