Poj 2887-Big String Splay

Posted 2020-06-22 微弱的世界

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。。。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cmath>
  5 #include<iostream>
  6 #include<algorithm>
  7 using namespace std;
  8 #define MAXN 1002010
  9 struct node
 10 {
 11     int left,right,size;
 12     char val;
 13 }tree[MAXN];
 14 int father[MAXN];
 15 char str[MAXN];
 16 void Pushup(int k)
 17 {
 18     tree[k].size=tree[tree[k].left].size+tree[tree[k].right].size+1;
 19 }
 20 void rotate(int x,int &root)
 21 {
 22     int y=father[x],z=father[y];
 23     if(y==root)root=x;
 24     else
 25     {
 26         if(tree[z].left==y)tree[z].left=x;
 27         else tree[z].right=x;
 28     }
 29     if(tree[y].left==x)
 30     {
 31         father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y;
 32     }
 33     else
 34     {
 35         father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y;
 36     }
 37     Pushup(y);Pushup(x);
 38 }
 39 void Splay(int x,int &root)
 40 {
 41     while(x!=root)
 42     {
 43         int y=father[x],z=father[y];
 44         if(y!=root)
 45         {
 46             if((tree[y].left==x)^(tree[z].left==y))rotate(x,root);
 47             else rotate(y,root);
 48         }
 49         rotate(x,root);
 50     }
 51 }
 52 void Build(int l,int r,int f)
 53 {
 54     if(l>r)return;
 55     int now=l,fa=f;
 56     if(l==r)
 57     {
 58         tree[now].val=str[l];tree[now].size=1;
 59         father[now]=fa;
 60         if(l<f)tree[fa].left=now;
 61         else tree[fa].right=now;
 62         return;
 63     }
 64     int mid=(l+r)/2;
 65     now=mid;
 66     Build(l,mid-1,mid);Build(mid+1,r,mid);
 67     tree[now].val=str[mid];father[now]=fa;
 68     Pushup(now);
 69     if(mid<f)tree[fa].left=now;
 70     else tree[fa].right=now;
 71 }
 72 int Find(int root,int rank)
 73 {
 74     if(rank==tree[tree[root].left].size+1)return root;
 75     if(rank<=tree[tree[root].left].size)return Find(tree[root].left,rank);
 76     else return Find(tree[root].right,rank-tree[tree[root].left].size-1);
 77 }
 78 int main()
 79 {
 80     int m,i,k,L,R,x,y,z,lstr,rt,n;
 81     char ch,fh[2];
 82     scanf("%s",str+2);
 83     lstr=strlen(str+2);
 84     m=lstr+2;
 85     Build(1,m,0);
 86     rt=(1+m)/2;
 87     scanf("%d",&n);
 88     for(i=1;i<=n;i++)
 89     {
 90         scanf("\n%s",fh);
 91         if(fh[0]==‘Q‘)
 92         {
 93             scanf("%d",&k);
 94             L=k;R=k+2;
 95             x=Find(rt,L);y=Find(rt,R);
 96             Splay(x,rt);Splay(y,tree[x].right);
 97             z=tree[y].left;
 98             printf("%c\n",tree[z].val);
 99         }
100         else
101         {
102             scanf("\n%c %d",&ch,&k);
103             L=k;R=k+1;
104             x=Find(rt,L);y=Find(rt,R);
105             Splay(x,rt);Splay(y,tree[x].right);
106             tree[y].left=++m;
107             z=tree[y].left;tree[z].size=1;
108             father[z]=y;tree[z].val=ch;
109             Pushup(y);Pushup(x);
110         }
111     }
112     return 0;
113 }

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