Query
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2859 Accepted Submission(s): 925
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
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题目大意：两种操作，1 2，操作为1时输入 a b c ,第a个字符串的第b位变成c，操作为2时，输入a ，询问从第a位起两个字符串有多少位是相等的
ac代码
#include<stdio.h>
#include<string.h>
#define min(a,b) (a>b?b:a)
int node[1000010<<2];
char s1[1000010],s2[1000010];
void pushup(int tr)
{
node[tr]=node[tr<<1]+node[tr<<1|1];
}
void build_tr(int l,int r,int tr)
{
if(l==r)
{
if(s1[l]==s2[l])
{
node[tr]=1;
}
else
node[tr]=0;
return;
}
int mid=(l+r)>>1;
build_tr(l,mid,tr<<1);
build_tr(mid+1,r,tr<<1|1);
pushup(tr);
}
void update(int pos,int l,int r,int tr)
{
if(l==r)
{
if(s1[l]==s2[l])
{
node[tr]=1;
}
else
node[tr]=0;
return;
}
int mid=(l+r)>>1;
if(pos<=mid)
{
update(pos,l,mid,tr<<1);
}
else
update(pos,mid+1,r,tr<<1|1);
pushup(tr);
}
int query(int pos,int l,int r,int tr)
{
if(l==r)
return node[tr];
if(pos>=l&&pos<=r)
{
if(node[tr]==r-l+1)
{
return r-pos+1;
}
else
{
int mid=(l+r)>>1;
int ans=0;
if(pos<=mid)
{
ans+=query(pos,l,mid,tr<<1);
if(ans==mid-pos+1)
ans+=query(mid+1,mid+1,r,tr<<1|1);
}
else
ans+=query(pos,mid+1,r,tr<<1|1);
return ans;
}
}
}
int main()
{
int t,c=0;
scanf("%d",&t);
while(t--)
{
scanf("%s%s",s1+1,s2+1);
int len1=strlen(s1+1);
int len2=strlen(s2+1);
int len=min(len1,len2);
build_tr(1,len,1);
int q;
scanf("%d",&q);
printf("Case %d:\n",++c);
while(q--)
{
int op;
scanf("%d",&op);
if(op==1)
{
char s[2];
int a,b;
scanf("%d%d%s",&a,&b,s);
if(a==1)
s1[b+1]=s[0];
else
s2[b+1]=s[0];
update(b+1,1,len,1);
}
else
{
int a;
scanf("%d",&a);
printf("%d\n",query(a+1,1,len,1));
}
}
}
}
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