BZOJ2212: [Poi2011]Tree Rotations(线段树合并)

题意

题目链接

sol

线段树合并为什么我会在这个时候学这种东西

就是暴力合并两棵线段树(必须动态开节点)，遇到空节点就返回

可以证明，对于\(m\)个仅有一个元素，权值范围在\([1, n]\)的线段树合并的复杂度为\(mlogn\)

对于此题来说，显然子树内与子树外互不影响，因此暴力判断一下翻转之后会不会变优就行了

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int maxn = 1e7 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, val[maxn], ls[maxn], rs[maxn], tot;
ll ans, c1, c2;
int newtree(int l, int r, int x) {
    val[++tot] = 1;
    if(l == r) return tot;
    int mid = l + r >> 1, now = tot;
    if(x <= mid) ls[now] = newtree(l, mid, x);
    else rs[now] = newtree(mid + 1, r, x);
    return now;
}
int merge(int l, int r, int x, int y) {
    if(!x || !y) return x + y;
    if(l == r) {val[++tot] = val[x] + val[y]; return tot;}
    c1 += 1ll * val[rs[x]] * val[ls[y]], c2 += 1ll * val[ls[x]] * val[rs[y]];
    int mid = l + r >> 1, now = ++tot;
    ls[now] = merge(l, mid, ls[x], ls[y]);
    rs[now] = merge(mid + 1, r, rs[x], rs[y]);
    val[now] = val[x] + val[y];
    return now;
}
int dfs() {
    int v = read();
    if(v) return newtree(1, n, v);
    int now = merge(1, n, dfs(), dfs());
    ans += min(c1, c2); c1 = c2 = 0;
    return now;
}
int main() {
    n = read();
    dfs();
    cout << ans;
    return 0;
}