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 [align=left]Problem Description[/align] 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]. [align=left]Input[/align] 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. [align=left]Output[/align] 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. [align=left]Sample Input[/align] 1aaabbaaabbaa 72 0 212 2 2 2 31 1b2 02 3 [align=left]Sample Output[/align] Case 1:21 10 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;
}