HDU 4339 Query【线段树】单点更新，动态查询

Query

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3010 Accepted Submission(s): 964

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
21
2 2
 2 
2 3
1 
1b
2 0
2 3

Sample Output

Case 1:
2
1
1
0
4
1

本来还感觉自己学的还可以，结果发现自己连递归都控制不好，一直出错，数组下标都弄混了

最后借鉴学长的思路才解决，自己还是迷迷糊糊的..........

找从某个位置处开始两个字符串连续相同的字符的个数........

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[2][1000005];int sum[4000005];void build(int rt,int l,int r)//建树没什么说的{if(l==r){sum[rt]=0;if(s[0][r]==s[1][r]){sum[rt]=1;}return;}int mid=(l+r)>>1,tp=rt<<1;build(tp,l,mid);build(tp|1,mid+1,r);sum[rt]=sum[tp]+sum[tp|1];}void update(int rt,int l,int r,int i)//更新也没什么说的{if(l==r){sum[rt]=0;//醉了，卡死在这了，坐标都混了 if(s[0][l]==s[1][l]){sum[rt]=1;}return;}int mid=(l+r)>>1,tp=rt<<1;if(i<=mid){update(tp,l,mid,i);}else{update(tp|1,mid+1,r,i);}sum[rt]=sum[tp]+sum[tp|1];}int find(int rt,int l,int r,int b){if(l==r)//单个元素的时候直接返回 {return sum[rt];}if(b>=l&&b<=r){if(sum[rt]==r-l+1)//一段都相同，直接返回{return r-b+1;}else{int mid=(l+r)>>1,tp=rt<<1,res=0;if(b<=mid)//查询点在中间点左边，需要讨论{res+=find(tp,l,mid,b);//先递归计算左半边有效区域if(res==mid-b+1)//左边全部为1{res+=find(tp|1,mid+1,r,mid+1);//则查3右边}}else{res+=find(tp|1,mid+1,r,b);//否则直接递归右边}return res;}}}int main(){int t,n,m;//freopen("shuju.txt","r",stdin);scanf("%d",&t);for(int k=1;k<=t;++k){memset(s,0,sizeof(s));scanf("%s%s%d",s[0]+1,s[1]+1,&m);n=min(strlen(s[0]+1),strlen(s[1]+1)); build(1,1,n);printf("Case %d:\n",k);while(m--){int kase;scanf("%d",&kase);if(kase==1){int a,i;char c[2];scanf("%d%d%s",&a,&i,c);s[a-1][i+1]=c[0];//修改单点 update(1,1,n,i+1);//更新树 }else{int b;scanf("%d",&b);printf("%d\n",find(1,1,n,b+1));//}}}return 0;}