P3521 [POI2011]ROT-Tree Rotations
题目大意：
给一棵((1≤n≤200000))个叶子的二叉树，可以交换每个点的左右子树，要求前序遍历叶子的逆序对最少。
我们发现交换两个子树并不会影响某个子树内的逆序对个数,只会对两个子树之间的逆序对产生影响.
所以我们将换与不换的逆序对求出来,取个最小值
将两颗线段树合并就好了
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cctype>
#include<algorithm>
#define LL long long
using namespace std;
const int N = 4e5 + 3;
struct node{
int sum;
int lc,rc;
}a[N * 30];
int root;
int n,t,cnt = 1;
int rt[N];
LL ans1,ans2,ans;
inline int read(){
int v = 0,c = 1;char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') c = -1;
ch = getchar();
}
while(isdigit(ch)){
v = v * 10 + ch - 48;
ch = getchar();
}
return v * c;
}
inline void pushup(int u){
a[u].sum = a[a[u].lc].sum + a[a[u].rc].sum;
}
inline void ins(int &u,int l,int r,int x){
// printf("%d %d %d
",l,r,x);
if(!u) u = ++t;
if(l == r){
a[u].sum++;
return ;
}
int mid = (l + r) >> 1;
if(x <= mid) ins(a[u].lc,l,mid,x);
else ins(a[u].rc,mid + 1,r,x);
pushup(u);
}
inline void merge(int &u1,int u2,int l,int r){
if(!u1){u1 = u2;return ;}
if(!u2) return ;
if(l == r){
a[u1].sum += a[u2].sum;
return ;
}
int mid = (l + r) >> 1;
ans1 += 1ll * a[a[u1].lc].sum * a[a[u2].rc].sum;
ans2 += 1ll * a[a[u2].lc].sum * a[a[u1].rc].sum;
merge(a[u1].lc,a[u2].lc,l,mid);
merge(a[u1].rc,a[u2].rc,mid + 1,r);
pushup(u1);
}
inline void solve(int pos){
int x = read();
if(!x){
int lc = ++cnt;solve(lc);
int rc = ++cnt;solve(rc);
merge(rt[pos],rt[lc],1,n);
ans1 = ans2 = 0;
merge(rt[pos],rt[rc],1,n);
ans += min(ans1,ans2);
}
else{
ins(rt[pos],1,n,x);
return ;
}
}
int main(){
n = read();
solve(1);
printf("%lld
",ans);
return 0;
}