Poj 2887-Big String Splay

题目：http://poj.org/problem?id=2887 Big String Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 6527   Accepted: 1563 Description You are given a string and supposed to do some string manipulations. Input The first line of the input contains the initial string. You can assume that it is non-empty and its length does not exceed 1,000,000. The second line contains the number of manipulation commands N (0 < N ≤ 2,000). The following N lines describe a command each. The commands are in one of the two formats below: I ch p: Insert a character ch before the p-th character of the current string. If p is larger than the length of the string, the character is appended to the end of the string. Q p: Query the p-th character of the current string. The input ensures that the p-th character exists. All characters in the input are digits or lowercase letters of the English alphabet. Output For each Q command output one line containing only the single character queried. Sample Input ab 7 Q 1 I c 2 I d 4 I e 2 Q 5 I f 1 Q 3 Sample Output a d e Source POJ Monthly--2006.07.30, zhucheng

题意：给一个字符串，要求查询某一位的字母，或在某一位前插入一个字母.

题解：

Splay的单点插入，和单点查询.

记得内存大小要把操作的2000加上。

速度422ms，还不错。。。

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define MAXN 1002010
 struct node
 {
     int left,right,size;
     char val;
 }tree[MAXN];
 int father[MAXN];
 char str[MAXN];
 void Pushup(int k)
 {
     tree[k].size=tree[tree[k].left].size+tree[tree[k].right].size+;
 }
 void rotate(int x,int &root)
 {
     int y=father[x],z=father[y];
     if(y==root)root=x;
     else
     {
         if(tree[z].left==y)tree[z].left=x;
         else tree[z].right=x;
     }
     if(tree[y].left==x)
     {
         father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y;
     }
     else
     {
         father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y;
     }
     Pushup(y);Pushup(x);
 }
 void Splay(int x,int &root)
 {
     while(x!=root)
     {
         int y=father[x],z=father[y];
         if(y!=root)
         {
             if((tree[y].left==x)^(tree[z].left==y))rotate(x,root);
             else rotate(y,root);
         }
         rotate(x,root);
     }
 }
 void Build(int l,int r,int f)
 {
     if(l>r)return;
     int now=l,fa=f;
     if(l==r)
     {
         tree[now].val=str[l];tree[now].size=;
         father[now]=fa;
         if(l<f)tree[fa].left=now;
         else tree[fa].right=now;
         return;
     }
     int mid=(l+r)/;
     now=mid;
     Build(l,mid-,mid);Build(mid+,r,mid);
     tree[now].val=str[mid];father[now]=fa;
     Pushup(now);
     if(mid<f)tree[fa].left=now;
     else tree[fa].right=now;
 }
 int Find(int root,int rank)
 {
     if(rank==tree[tree[root].left].size+)return root;
     if(rank<=tree[tree[root].left].size)return Find(tree[root].left,rank);
     else return Find(tree[root].right,rank-tree[tree[root].left].size-);
 }
 int main()
 {
     int m,i,k,L,R,x,y,z,lstr,rt,n;
     char ch,fh[];
     scanf("%s",str+);
     lstr=strlen(str+);
     m=lstr+;
     Build(,m,);
     rt=(+m)/;
     scanf("%d",&n);
     for(i=;i<=n;i++)
     {
         scanf("\n%s",fh);
         if(fh[]=='Q')
         {
             scanf("%d",&k);
             L=k;R=k+;
             x=Find(rt,L);y=Find(rt,R);
             Splay(x,rt);Splay(y,tree[x].right);
             z=tree[y].left;
             printf("%c\n",tree[z].val);
         }
         else
         {
             scanf("\n%c %d",&ch,&k);
             L=k;R=k+;
             x=Find(rt,L);y=Find(rt,R);
             Splay(x,rt);Splay(y,tree[x].right);
             tree[y].left=++m;
             z=tree[y].left;tree[z].size=;
             father[z]=y;tree[z].val=ch;
             Pushup(y);Pushup(x);
         }
     }
     return ;
 }