解题关键:线段树合并模板题.线段树合并的题目一般都是权值线段树,因为结构相同,求逆序对时,遍历权值线段树的过程就是遍历所有mid的过程,所有能求出所有逆序对. #include<iostream> #include<cstdio> #define ll long long using namespace std; int n,sz,seg; ll ans,cnt1,cnt2; ],l[],r[],root[]; ],ls[],rs[]; //动态开点线段树 //int new_no…
[BZOJ2212][Poi2011]Tree Rotations Description Byteasar the gardener is growing a rare tree called Rotatus Informatikus. It has some interesting features: The tree consists of straight branches, bifurcations and leaves. The trunk stemming from the gro…
Description Byteasar the gardener is growing a rare tree called Rotatus Informatikus. It has some interesting features: The tree consists of straight branches, bifurcations and leaves. The trunk stemming from the ground is also a branch. Each branch…
BZOJ_2212_[Poi2011]Tree Rotations_线段树合并 Description Byteasar the gardener is growing a rare tree called Rotatus Informatikus. It has some interesting features: The tree consists of straight branches, bifurcations and leaves. The trunk stemming from t…
原文地址:http://www.cnblogs.com/GXZlegend/p/6826614.html 题目描述 Byteasar the gardener is growing a rare tree called Rotatus Informatikus. It has some interesting features: The tree consists of straight branches, bifurcations and leaves. The trunk stemming…
2212: [Poi2011]Tree Rotations Time Limit: 20 Sec Memory Limit: 259 MB Description Byteasar the gardener is growing a rare tree called Rotatus Informatikus. It has some interesting features: The tree consists of straight branches, bifurcations and lea…
题目链接:https://www.lydsy.com/JudgeOnline/problem.php?id=2212 思路:用线段树合并求出交换左右儿子之前之后逆序对的数量,如果数量变小则交换. 实现代码: #include<bits/stdc++.h> using namespace std; #define ll long long ; int n,cnt,idx; ll ans,cnt1,cnt2; int v[M],l[M],r[M],root[M]; ],ls[M*],rs[M*];…