UVALive 6145 Version Controlled IDE（可持久化treap、rope）

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4156

题目拷贝难度大我就不复制了。

题目大意：维护一个字符串，要求支持插入、删除操作，还有输出第 i 次操作后的某个子串。强制在线。

思路1：使用可持久化treap可破，详细可见CLJ的《可持久化数据结构的研究》。

思路2：rope大法好，详见：http://blog.csdn.net/guognib/article/details/20563453（文档：http://www.sgi.com/tech/stl/Rope.html），代码短速度快，可惜不能打lazy标记。

PS：自从学了函数式编程，发现可持久化什么的都变简单了。

PS：不用智能指针只要845MS。这真是一个大坑。本来我换成普通指针只是想用于调试……

PS：UVALive居然不保存代码！于是我又去vjudge交了一次。

代码（C++11 2116MS）：

 #include <bits/stdc++.h>
 using namespace std;

 struct Node;
 typedef shared_ptr<Node> Nptr;
 //typedef Node* Nptr;

 struct Node {
     Nptr lson, rson;
     int size, weight;
     char c;
     void update() {
         size = lson->size + rson->size + ;
     }
 };
 Nptr nil;

 Nptr new_node(char c) {
     Nptr x = Nptr(new Node);
     x->lson = x->rson = nil;
     x->size = ;
     x->weight = rand();
     x->c = c;
     return x;
 }

 Nptr new_node(char c, Nptr lson, Nptr rson) {
     Nptr x = Nptr(new Node);
     x->lson = lson;
     x->rson = rson;
     x->update();
     x->weight = rand();
     x->c = c;
     return x;
 }

 void initTreap() {
     nil = Nptr(new Node);
     nil->lson = nil->rson = nil;
     nil->size = ;
 }

 Nptr build(char *st, char *ed) {
     if(st == ed) return nil;
     assert(st < ed);
     char *mid = st + (ed - st) / ;
     return new_node(*mid, build(st, mid), build(mid + , ed));
 }

 typedef pair<Nptr, Nptr> Ppp;

 Ppp split(Nptr x, int n) {
     if(n == ) return make_pair(nil, x);
     int ls = x->lson->size;

     if(ls >= n) {
         Ppp a = split(x->lson, n);
         return make_pair(a.first, new_node(x->c, a.second, x->rson));
     } else {
         Ppp a = split(x->rson, n - ls - );
         return make_pair(new_node(x->c, x->lson, a.first), a.second);
     }
 }

 Nptr merge(Nptr a, Nptr b) {
     if(a == nil) return b;
     if(b == nil) return a;
     if(a->weight < b->weight) {
         return new_node(a->c, a->lson, merge(a->rson, b));
     } else {
         return new_node(b->c, merge(a, b->lson), b->rson);
     }
 }

 int print(Nptr x) {
     if(x == nil) return ;
     int res = (x->c == 'c');
     res += print(x->lson);
     putchar(x->c);
     res += print(x->rson);
     return res;
 }

 Nptr insert(Nptr x, int pos, char s[]) {
     Nptr a = build(s, s + strlen(s));
     Ppp p = split(x, pos);
     return merge(p.first, merge(a, p.second));
 }

 Nptr remove(Nptr x, int pos, int len) {
     Ppp a = split(x, pos);
     Ppp b = split(a.second, len);
     return merge(a.first, b.second);
 }

 int print(Nptr x, int pos, int len) {
     Ppp a = split(x, pos);
     Ppp b = split(a.second, len);
     int res = print(b.first);
     puts("");
     return res;
 }

 Nptr rt[];
 char s[];
 int n, d, vnow;

 int main() {
     initTreap();
     rt[] = nil;

     scanf("%d", &n);
     while(n--) {
         int v, p, c, op;
         scanf("%d", &op);
         if(op == ) {
             scanf("%d%s", &p, s);
             p -= d;
             vnow++;
             rt[vnow] = insert(rt[vnow - ], p, s);
         }
         if(op == ) {
             scanf("%d%d", &p, &c);
             p -= d, c -= d;
             vnow++;
             rt[vnow] = remove(rt[vnow - ], p - , c);
         }
         if(op == ) {
             scanf("%d%d%d", &v, &p, &c);
             v -= d, p -= d, c -= d;
             d += print(rt[v], p - , c);
         }
     }
 }

代码（rope大法 322MS）：

 #include <bits/stdc++.h>
 #include <ext/rope>
 using namespace std;
 #define FOR(i, n) for(int i = 0; i < n; ++i)

 const int MAXN = ;
 const int MAXS = ;

 __gnu_cxx::crope rt[MAXN], tmp;

 char s[MAXS];
 int m, vnow, d;

 int main() {
     scanf("%d", &m);
     while(m--) {
         int op, p, v, c;
         scanf("%d", &op);
         if(op == ) {
             scanf("%d%s", &p, s);
             p -= d;
             rt[vnow + ] = rt[vnow];
             rt[++vnow].insert(p, s);
         } else if(op == ) {
             scanf("%d%d", &p, &c);
             p -= d, c -= d;
             rt[vnow + ] = rt[vnow];
             rt[++vnow].erase(p - , c);
         } else if(op == ) {
             scanf("%d%d%d", &v, &p, &c);
             v -= d, p -= d, c -= d;
             tmp = rt[v].substr(p - , c);
             printf("%s\n", tmp.c_str());
             d += count(tmp.begin(), tmp.end(), 'c');
         }
     }
 }