BZOJ1269——[AHOI2006]文本编辑器editor

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1、题意：各种splay操作，一道好的模板题2333

2、分析：splay模板题，没啥解释QAQ

#include <stack> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 2000010
 
inline int read(){
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9'){
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
 
namespace splay{
    struct Node{
        Node *ch[2], *fa;
        char c;
        bool rev;
        int size;
         
        inline int which();
         
        inline void reverse(){
            rev ^= 1;
            swap(ch[0], ch[1]);
        } 
         
        inline void pd(){
            if(rev){
                ch[0] -> reverse();
                ch[1] -> reverse();
                rev = false; 
            }
        }
         
        inline void maintain(){
            size = 1 + ch[0] -> size + ch[1] -> size;
        }
         
        Node();
    } *null = new Node, ft[M];
     
    int tot;
     
    Node::Node(){
        size = 1;
        c = '\0';
        ch[0] = ch[1] = fa = null;
        rev = false;
    } 
     
    inline int Node::which(){
        if(fa == null) return -1;
        return this == fa -> ch[1];
    }
     
    inline void rotate(Node *o){
        Node *p = o -> fa;
        int l = o -> which(), r = l ^ 1;
        o -> fa = p -> fa;
        if(p -> which() != -1) p -> fa -> ch[p -> which()] = o;
        p -> ch[l] = o -> ch[r];
        if(o -> ch[r]) o -> ch[r] -> fa = p;
        o -> ch[r] = p; p -> fa = o;
        o -> ch[r] -> maintain();
        o -> maintain();
    }
     
    inline void splay(Node *o){
        static stack<Node*> st;
        if(!o) return;
        Node *p = o;
        while(1){
            st.push(p);
            if(p -> which() == -1) break;
            p = p -> fa;
        }
        while(!st.empty()){
            st.top() -> pd();
            st.pop();
        }
        while(o -> which() != -1){
            p = o -> fa;
            if(p -> which() != -1){
                if(p -> which() ^ o -> which()) rotate(o);
                else rotate(p);
            }
            rotate(o);
        }
    }
     
    inline Node* Kth(Node *o, int k){
        o -> pd(); 
        int t = o -> ch[0] -> size + 1;
        if(k == t) return o;
        if(k < t) return Kth(o -> ch[0], k);
        return Kth(o -> ch[1], k - t);
    }
     
    inline Node *merge(Node *a, Node *b){
        if(a == null) return b;
        if(b == null) return a;
        Node *p = Kth(a, a -> size);
        splay(p);
        Node *c = p;
        c -> ch[1] = b;
        b -> fa = c;
        c -> maintain();
        return c;
    }
     
    inline void split(Node *o, int k, Node* &a, Node* &b){
        if(k == 0){
            a = null;
            b = o;
            return;
        }
        if(k == o -> size){
            a = o;
            b = null;
            return;
        }
        Node *p = Kth(o, k);
        splay(p);
        o = p;
        o -> maintain();
        b = p -> ch[1];
        b -> fa = null;
        p -> ch[1] = null;
        a = p;
        a -> maintain();
    }
} 
 
using namespace splay;
 
int main(){
    null -> ch[0] = null -> ch[1] = null -> fa = null;
    null -> c = '\0'; null -> rev = false; null -> size = 0;
    int n = read(); int now = 0;
    Node *root = null; 
    char str[10]; 
    for(int i = 1; i <= n; i ++){
        scanf("%s", str);
        if(str[0] == 'M'){
            int x = read(); now = x;
        } 
        else if(str[0] == 'I'){
            int x = read();
            Node *a, *b, *d = null;
            split(root, now, a, b);
            for(int i = 1; i <= x; i ++){
                char ch = getchar();
                Node *c = &ft[++ tot];
                c -> c = ch;
                d = merge(d, c);
            }
            root = merge(a, d);
            root = merge(root, b);
        }
        else if(str[0] == 'D'){
            int x = read();
            Node *a, *b, *c;
            split(root, now, a, b);
            split(b, x, b, c);
            root = merge(a, c);
        }
        else if(str[0] == 'R'){
            int x = read();
            Node *a, *b, *c;
            split(root, now, a, b);
            split(b, x, b, c);
            b -> reverse();
            root = merge(a, b);
            root = merge(root, c);
        }
        else if(str[0] == 'G'){
            Node *p = Kth(root, now + 1);
            printf("%c\n", p -> c);
        }
        else if(str[0] == 'P'){
            now --;
        }
        else{
            now ++;
        }
    }
    return 0;
}

BZOJ1269——[AHOI2006]文本编辑器editor

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原文地址：http://blog.csdn.net/qzh_1430586275/article/details/51893210