hdu4339 Query　树状数组

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线段树&&树状数组

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本文深入探讨AI音视频处理领域的关键技术，重点介绍视频分割与语义识别，通过实例解析如何利用AI技术实现复杂场景下的视频内容理解与应用。

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Query

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2087 Accepted Submission(s): 720

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
树状数组就是好啊，这题把两个串中相等的化成１不等的化成０，用树状数组，就可以快速查出从i到j的和，如果这个和等于区间的宽度，不就说明，是相等的串么？我们还可以用二分更快的查出，但这题，同样的原理，用线段树就是过不了！看来线段树常数，还是太大了啊！
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAXN 1000005
char str[2][MAXN];
int prime[MAXN];
int l[MAXN];
int all,len;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int c)
{
    int i;

    while(x<MAXN)
    {
        l[x]+=c;
        x=x+lowbit(x);
    }
}
int query(int b)
{
    int i,sum=0,x=b;
    while(x>0)
    {
        sum+=l[x];
        x=x-lowbit(x);
    }
    return sum;
}
int  twocut(int x)
{
    int s,e,mid,num;
    x++;
    s=x;e=len;
    num=s-1;
    while(s<=e)//kdfljdkfljsdlkf
    {
        mid=(s+e)>>1;
       if(query(mid)-query(num)==mid-num)//·?o?
        {
            s=mid+1;
        }
        else
        {
            e=mid-1;
        }
    }
    return e-num;
}
int fmin(int a,int b)
{
    if(a<b)
    return a;
    return b;
}
int main()
{
   int tcase,tt,i,asknum,tempask,strnum,index;
   char c;
   scanf("%d",&tcase);
   for(tt=1;tt<=tcase;tt++)
   {
       memset(l,0,sizeof(l));
       memset(prime,0,sizeof(prime));
       printf("Case %d:\n",tt);
       getchar();
       gets(str[0]);
       //getchar();
       gets(str[1]);
       len=fmin(strlen(str[0]),strlen(str[1]));
       //printf("%d\n",len);
       for(i=0;i<len;i++)
       {
           if(str[0][i]!=str[1][i])
           {
               prime[i]=0;
           }
           else
           {
               prime[i]=1;
               update(i+1,1);
           }
        }
       all=0;
       scanf("%d",&asknum);
       while(asknum--)
       {
           scanf("%d",&tempask);
           if(tempask==1)
           {
               scanf("%d%d %c",&strnum,&index,&c);
               strnum--;//
               str[strnum][index]=c;
               if(c!=str[strnum^1][index]&&prime[index])
               {
                   prime[index]=0;
                   update(index+1,-1);

               }
               else if(c==str[strnum^1][index]&&!prime[index])
               {
                   prime[index]=1;
                   update(index+1,1);

               }
           }
           else
           {
               scanf("%d",&index);

               printf("%d\n",twocut(index));
           }
       }
   }
    return 0;
}