Query on a string（线段树）（2017-icpc-乌鲁木齐网络赛）

                                                                                                

You have two strings $S$ and $T$ in all capitals.

Now an efficient program is required to maintain a operation and support a query.

The operation $Cich$ with given integer $i$ and capital letter $ch$, changes the $i$-th character of $S$ into $ch$.

The query $Qij$ asks the program to find out, in the substring of $S$ from the $i$-th character to the $j$-th one, the total number of $T$ appearing.

Input Format

The first line contains an integer $T$, indicating that there are $T$ test cases.

For each test case, the first line contains an integer $N(N\le 100000)$.

The second line is the string $S(|S|\le 100000)$ and the third line is the string $T(|T|\le 10)$.

Each of the following $N$ lines provide a operation or a query as above descriptions.

Output Format

For each query, output an integer correponding to the answer.

Output an empty line after each test case.

样例输入

1
5
AABBABA
AA
Q 1 3
C 6 A
Q 2 7
C 2 B
Q 1 5

样例输出

1
2
0

题意：t组测试数据，n次操作，给两个字符串S，T；，Q  X   Y表示查询在S串中第X--Y这段字符串内T串出现了几次；C   X  ch  将S串中第X个字符改成ch

思路：将第一个串处理一下，如果以当前字符为第一个字符，若出现了第二个字符串，用另一个数组将当前点记录为1，否则为0；如样例AABBABA   AA，记录的数组为1000000，然后以这个记录数组建线段树，查询就是求这个区间的和，更改就是线段树单点更新，但是要注意S字符串最后几个到最后长度都比T字符串小就不用判断了；具体看代码：

AC代码：

#include
#include
#include
#include
#include
#define maxn 100005
using namespace std;
char str[maxn],s[maxn];
int arr[maxn],len,ll;
void Getarr(char *str,char *s)
{
    for(int i=0;ilen)//如果i+ll>len，说明到最后都不能构成T
            {
                flag=0;
                arr[i]=0;
                break;
            }
        }
        if(flag)
            arr[i]=1;
    }
}
struct node
{
    int l;
    int r;
    int sum;
}segtree[maxn*4];
void pushup(int root)
{
    segtree[root].sum=segtree[root<<1].sum+segtree[root<<1|1].sum;
}
void build(int root,int l,int r)
{
    segtree[root].l=l;
    segtree[root].r=r;
    if(l==r)
    {
         segtree[root].sum=arr[l];
         return;
    }
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
    pushup(root);
}
void update(int root,int p,int val)
{
    int ll=segtree[root].l;
    int rr=segtree[root].r;
    if(ll==rr)
    {
        segtree[root].sum=val;
        return;
    }
    int mid=(ll+rr)>>1;
    if(p<=mid) update(root<<1,p,val);
    else update(root<<1|1,p,val);
    pushup(root);
}
int query(int root,int l,int r)
{
    int ll=segtree[root].l;
    int rr=segtree[root].r;
    if(l<=ll&&rr<=r)
        return segtree[root].sum;
    int mid=(ll+rr)>>1;
    int ans=0;
    if(l<=mid)
        ans+=query(root<<1,l,r);
    if(r>mid)
        ans+=query(root<<1|1,l,r);
    return ans;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%s%s",&n,str,s);
        len=strlen(str);ll=strlen(s);
        Getarr(str,s);
        build(1,0,len-1);
        while(n--)
        {
            char op[5],ch[3];
            int x,y;
            scanf("%s",op);
            if(op[0]=='Q')
            {
                scanf("%d%d",&x,&y);
                if(y-x+1=0?x-ll+1:0;
                for(int i=tmp;i<=x;i++)
                {
                    int flag=1;
                    for(int j=0;jlen)
                        {
                            if(arr[i]!=0)
                                update(1,i,0);
                            arr[i]=0;
                            flag=0;
                            break;
                        }
                    }
                    if(flag==1)
                    {
                        if(arr[i]!=1)
                            update(1,i,1);
                        arr[i]=1;
                    }
                }
            }

        }
        cout<