hdu 4339 Query

Query

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1009    Accepted Submission(s): 302

Problem Description

You are given two strings s1[0..l1], s2[0..l2] and Q - number of queries.
Your task is to answer next queries:
  1) 1 a i c - you should set i-th character in a-th string to c;
  2) 2 i - you should output the greatest j such that for all k (i<=k and k<i+j) s1[k] equals s2[k].

 

Input

The first line contains T - number of test cases (T<=25).
Next T blocks contain each test.
The first line of test contains s1.
The second line of test contains s2.
The third line of test contains Q.
Next Q lines of test contain each query:
  1) 1 a i c (a is 1 or 2, 0<=i, i<length of a-th string, 'a'<=c, c<='z')
  2) 2 i (0<=i, i<l1, i<l2)
All characters in strings are from 'a'..'z' (lowercase latin letters).
Q <= 100000.
l1, l2 <= 1000000.

 

Output

For each test output "Case t:" in a single line, where t is number of test (numbered from 1 to T).
Then for each query "2 i" output in single line one integer j.

 

Sample Input

1

aaabba

aabbaa

7

2 0

2 1

2 2

2 3

1 1 2 b

2 0

2 3

 

Sample Output

Case 1:

2

1

0

1

4

1

 

Source

2012 Multi-University Training Contest 4

 

Recommend

zhoujiaqi2010

//len代表区间最左开始最长公共子串
//然后  if(len[k<<1]==m-l+1) len[k]=len[k<<1]+len[k<<1|1]; else len[k]=len[k<<1];
//开心的今早上1Y，本来打算昨晚写的、结果去看爱情公寓3了、、^_^
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <vector>
#define N 1000000
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
using namespace std;
char s1[N+2],s2[N+2];

int len[N<<2];
void up(int &k,int &l,int &m)
{
    if(len[k<<1]==m-l+1)
     len[k]=len[k<<1]+len[k<<1|1];
    else
     len[k]=len[k<<1];
}
void build(int l,int r,int k)
{
    if(l==r)
    {
      if(s1[l]==s2[l])
         len[k]=1;
        else
         len[k]=0;
      return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    up(k,l,m);
}
void update(int &i,int l,int r,int k)
{
    if(l==r)
    {
       if(s1[l]==s2[l])
         len[k]=1;
        else
         len[k]=0;
      return;
    }
    int m=(l+r)>>1;
    if(i<=m) update(i,lson);
    else     update(i,rson);
    up(k,l,m);
}
int query(int &i,int l,int r,int k)
{
    if(l==r)
     { return len[k];}
    int m=(l+r)>>1,t;
    if(i<=m)
     {
        t=query(i,lson);
        if(t==m-i+1)
         return t+len[k<<1|1];
        return t;
     }
    return query(i,rson);
}
int main()
{
    int l,l1,l2;
    int t=1,T;
    int q;
    scanf("%d",&T);
    int op,a,i;
    char c;
    while(T--)
    {
        scanf("%s",s1);
        scanf("%s",s2);
        l1=strlen(s1);
        l2=strlen(s2);
        l=l1<l2?l1:l2;//选取短的串建树
        l--;          //因为下标从0开始
        build(0,l,1);
        scanf("%d",&q);
        printf("Case %d:\n",t++);
        while(q--)
        {
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d %d %c",&a,&i,&c);
                if(i>l) continue;//超过长度更新没意义
                if(a==1)s1[i]=c;
                else s2[i]=c;
                update(i,0,l,1);
            }
            else
            {
              scanf("%d",&i);
              if(i>l){printf("0\n");continue;}//超长的话肯定是0
              printf("%d\n",query(i,0,l,1));
            }
        }
    }
    return 0;
}