[UVA12538]Version Controlled IDE && 可持久化treap

似乎不是特别难写的样子？

然而我并不会rope

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
using namespace std;
typedef long long LL;
inline int read() {
	int x=0, f=1; char ch=getchar();
	while('0'>ch || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
	while('0'<=ch && ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}
const int MAXN = 7000000;
int q, ccnt;  
int now, root[50020];  
char str[120];
struct Treap {
	int val[MAXN+10], sz[MAXN+10], ch[MAXN+10][2], rnd[MAXN+10];
	int ncnt;
	void up(int x) {
		sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
	}
	int NewNode(int x) {
		int k = ++ncnt;
		ch[k][0] = ch[x][0]; ch[k][1] = ch[x][1];
		rnd[k] = rnd[x]; val[k] = val[x]; sz[k] = 1;
		return k;
	}
	int merge(int x, int y) {
		if(!x) return y;
		if(!y) return x;
		if(rnd[x] < rnd[y]) {
			int k = NewNode(x);
			ch[k][1] = merge(ch[x][1], y);
			up(k);
			return k;
		}
		else {
			int k = NewNode(y);
			ch[k][0] = merge(x, ch[y][0]);
			up(k);
			return k;
		}
	}
	void split(int x, int k, int &L, int &R) {
		if(sz[x] <= k) {
			L = x, R = 0;
			return ;
		}
		int tmp = sz[ch[x][0]], t = NewNode(x);
		int a, b;
		if(k <= tmp) {
			split(ch[x][0], k, a, b);
			ch[t][0] = b;
			R = t;
			L = a;
		}
		else {
			split(ch[x][1], k-tmp-1, a, b);
			ch[t][1] = a;
			L = t;
			R = b;
		}
		up(t);
	}
	int build(int l, int r, int pre) {
		if(l > r) return 0;
		int m = (l+r) >> 1;
		int x = ++ncnt;
		rnd[x] = rand()+pre;
		sz[x] = 1; val[x] = str[m];
		ch[x][0] = build(l, m-1, rnd[x]);
		ch[x][1] = build(m+1, r, rnd[x]);
		up(x);
		return x;
	}
	void out(int x) {
		if(!x) return ;
		out(ch[x][0]);
		putchar(val[x]); if(val[x] == 'c') ccnt++;
		out(ch[x][1]);
	}
	void solve() {
		ncnt = ccnt = now = 0;
		memset(root, 0, sizeof(root));
		for(int i = 1; i <= q; i++) {
			int op = read();
			if(op == 1) {
				int p = read()-ccnt;
				SF("%s", str+1);
				int t = build(1, strlen(str+1), 0);
				now++;
				if(!p) root[now] = merge(t, root[now-1]);
				else if(p == sz[root[now-1]]) root[now] = merge(root[now-1], t);
				else {
					int L, R;
					split(root[now-1], p, L, R);
					root[now] = merge(L, t);
					root[now] = merge(root[now], R);
				}
			}
			else if(op == 2) {
				now++;
				int p = read()-ccnt, c = read()-ccnt;
				int L, R, X, Y;
				split(root[now-1], p-1, L, R);
				split(R, c, X, Y);
				root[now] = merge(L, Y);
			}
			else {
				int v = read()-ccnt, p = read()-ccnt, c = read()-ccnt;
				int L, R, X, Y;
				split(root[v], p-1, L, R);
				split(R, c, X, Y);
				out(X);
				puts("");
			}
		}
	}
} tp;
int main() {
	while(~SF("%d", &q)) {
		tp.solve();
	}
}