Educational Codeforces Round 15 E 树上倍增 RMQ思想

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最新推荐文章于 2024-10-12 13:03:04 发布

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分类专栏： -----Codeforces----- -------数据结构------- RMQ | LCA | 倍增

本文链接：https://blog.youkuaiyun.com/libin66/article/details/52086069

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本文介绍了一种针对功能性图的路径分析算法，旨在解决给定长度k的路径上所有边的权重之和及其最小权重的问题。通过使用类似树上倍增的技术，文章详细阐述了如何高效地计算出从每个节点出发的指定长度路径的相关数据。

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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 1e9
using 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;
}