Educational Codeforces Round 15 E 树上倍增 RMQ思想

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E. Analysis of Pathes in Functional Graph

time limit per test2 seconds

memory limit per test512 megabytes

inputstandard input

outputstandard output

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 si, mi 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

题意：

n个点n条边的图，每个节点都连向一个点，可以是自己。

问从该点到长度为k的路径上。输出路径权值总和以及路径上的最小值

思路：

学习了一下树上倍增。其实理解了RMQ写这个还很快的。

fa[j][i]=fa[fa[j][i-1]][i-1]  这个很关键啊   表示以j节点开始到长度为i的路径上  所到的节点

代码：

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double lb;#define INF (1ll<<60)-1#define Max 1e9using namespace std;int n;int fa[100100][40];ll s[100100][40],m[100100][40],k;int main(){    scanf("%d%I64d",&n,&k);    for(int i=0;i<n;i++) scanf("%d",&fa[i][0]);    for(int i=0;i<n;i++){        scanf("%I64d",&m[i][0]);        s[i][0]=m[i][0];    }    for(int i=1;(1LL<<i)<=k;i++){        for(int j=0;j<n;j++){            fa[j][i]=fa[fa[j][i-1]][i-1];            m[j][i]=min(m[j][i-1],m[fa[j][i-1]][i-1]);            s[j][i]=s[j][i-1]+s[fa[j][i-1]][i-1];        }    }    for(int i=0;i<n;i++){        ll mn=INF,sum=0;        int x=i;        for(int j=0;(1LL<<j)<=k;j++){            if((1LL<<j)&k){                mn=min(mn,m[x][j]);                sum+=s[x][j];                x=fa[x][j];            }        }        printf("%I64d %I64d\n",sum,mn);    }    return 0;}