hdu4339 Query

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

这题属于区间合并，维护线段的llen（线段从左端点开始向右的最长连续1的长度），rlen（线段从右端点开始向左的最长连续1的长度），tlen（线段中最长连续1的长度，记录这个主要是为了剪枝，减少时间）。一开始先初始化，两个字符串，相同的部分记为1，不同的记为0，注意两个字符串的长度可能不同。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 1000100
char s1[maxn],s2[maxn];
int a[maxn],num;
struct node{
    int l,r,llen,rlen,tlen;
}b[4*maxn];

void pushup(int i)
{
    b[i].tlen=max(b[i*2].tlen,b[i*2+1].tlen);
    b[i].tlen=max(b[i].tlen,b[i*2].rlen+b[i*2+1].llen);

    b[i].llen=b[i*2].llen;b[i].rlen=b[i*2+1].rlen;
    if(b[i*2].llen==b[i*2].r-b[i*2].l+1)b[i].llen+=b[i*2+1].llen;
    if(b[i*2+1].rlen==b[i*2+1].r-b[i*2+1].l+1)b[i].rlen+=b[i*2].rlen;
}

void build(int l,int r,int i)
{
    int mid;
    b[i].l=l;b[i].r=r;
    if(l==r){
        b[i].tlen=b[i].llen=b[i].rlen=a[l];return;
    }
    mid=(l+r)/2;
    build(l,mid,i*2);
    build(mid+1,r,i*2+1);
    pushup(i);
}

void update(int id,int value,int i)
{
    int mid;
    if(b[i].l==b[i].r){
        b[i].tlen=b[i].llen=b[i].rlen=value;return;
    }
    mid=(b[i].l+b[i].r)/2;
    if(mid>=id)update(id,value,i*2);
    else update(id,value,i*2+1);
    pushup(i);
}

void question(int id,int i)
{
    int mid;
    if(b[i].l==b[i].r){
        num=1;return;
    }
    if(b[i].tlen==b[i].r-b[i].l+1){
        num=b[i].r-id+1;return;
    }
    mid=(b[i].l+b[i].r)/2;
    if(mid>=id){                //用4个剪枝试一试
        if(b[i*2].r-b[i*2].rlen+1<=id){
            num=b[i*2].r-id+1+b[i*2+1].llen;return;
        }
        else{
            question(id,i*2);
        }
    }
    else{
        if(b[i*2+1].l+b[i*2+1].llen-1>=id){
            num=b[i*2+1].l+b[i*2+1].llen-1-id+1;return;
        }
        else question(id,i*2+1);
    }
}

int main()
{
    int n,m,i,j,T,len1,len2,len,d,e,flag,c,num1=0;
    char f[10];
    scanf("%d",&T);
    while(T--)
    {
        num1++;
        printf("Case %d:\n",num1);
        scanf("%s%s",s1+1,s2+1);
        len1=strlen(s1+1);len2=strlen(s2+1);
        len=min(len1,len2);
        for(i=1;i<=len;i++){
            if(s1[i]==s2[i])a[i]=1;
            else a[i]=0;
        }
        build(1,len,1);
        scanf("%d",&m);
        for(i=1;i<=m;i++){
            scanf("%d",&c);
            if(c==1){
                scanf("%d%d%s",&d,&e,f);
                e++;
                if(e>len)continue;
                if(s1[e]==s2[e])flag=1;
                else flag=0;

                if(d==1)s1[e]=f[0];
                else s2[e]=f[0];
                if(s1[e]==s2[e] && flag==0)update(e,1,1);
                else if(s1[e]!=s2[e] && flag==1)update(e,0,1); //这里可以节省200ms
            }
            else if(c==2){
                scanf("%d",&d);
                d++;
                if(d>len || s1[d]!=s2[d]){
                    printf("0\n");continue;
                }
                num=0;
                question(d,1);
                printf("%d\n",num);
            }
        }
    }
}