HDU 4339 Query(线段树)

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

1aaabbaaabbaa72 02 12 22 31 1 2 b2 02 3

 

Sample Output

Case 1:210141

 

Source

2012 Multi-University Training Contest 4

 

并不难的线段树题目，然而比赛的时候没有做出来。。。准确的说是没有做。。。

首先是时间不够，看了一眼觉得区间合并会很麻烦，再加上线段树写得不是很溜，就没写

赛后仔细一看发现其实这题的区间合并蛮简单的。。。。。

保存区间[l,r]以l为起点的最长段为s

合并的时候只需要判断左子的s是否覆盖了左子的整个区间

若覆盖，则加上右子的s即可

另外，注意两个的串的长度没有说相等。。。

具体操作见代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<set>#include<stack>#include<cmath>#include<map>#include<stdlib.h>#include<cctype>#define mem(a,x) memset(a,x,sizeof(a))#define esp 1e-8using namespace std;typedef long long ll;const int N = 1000000;struct Node{    int l,r;    int s;//区间[l,r]内以l为起点的最长相等段}q[N*4];string a,b;void build(int i,int l,int r){    q[i].l = l;q[i].r = r;    if (l == r)    {        if (a[l] == b[l]) q[i].s = 1;        else q[i].s = 0;        return;    }    int mid = (l+r)>>1;    build(i<<1,l,mid);    build(i<<1|1,mid+1,r);    q[i].s = q[i<<1].s;//先等于左子的最大长度    if (q[i<<1].s == mid - l + 1)//如果左子的整个区间都相等    {        q[i].s += q[i<<1|1].s;//那么还可以加上右子    }}void update(int i,int x)//更改了第x个字符{    if (q[i].l == q[i].r)    {        if (a[q[i].l] == b[q[i].l]) q[i].s = 1;        else q[i].s = 0;        return ;    }    int mid = (q[i].l + q[i].r)>>1;    if (x <= mid) update(i<<1,x);    else update(i<<1|1,x);    q[i].s = q[i<<1].s;//先等于左子的最大长度    if (q[i<<1].s == mid - q[i].l + 1)//如果左子的整个区间都相等    {        q[i].s += q[i<<1|1].s;//那么还可以加上右子    }}int query(int i,int x){    if (q[i].l == q[i].r)    {        return q[i].s;    }    int mid = (q[i].l + q[i].r)>>1;    if (x <= mid)    {        int tmp = query(i<<1,x);        if (tmp == mid - x + 1) return tmp + q[i<<1|1].s;        else return tmp;    }    else return query(i<<1|1,x);}int main(){    int T;scanf("%d",&T);int kas = 0;    while (T--)    {        cin>>a>>b;        int la = a.size(),lb = b.size();        int n = min(la,lb)-1;        build(1,0,n);        int q;scanf("%d",&q);        printf("Case %d:\n",++kas);int op,x,i;char c;        while (q--)        {            scanf("%d",&op);            if (op == 1)            {                scanf("%d %d %c",&x,&i,&c);                if (i > n) continue;//越界的不用更新了                if (x == 1) a[i] = c;                else b[i] = c;                update(1,i);            }            else            {                scanf("%d",&i);                if (i > n) puts("0");                else printf("%d\n",query(1,i));            }        }    }    return 0;}

线段树还是不够熟练啊。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。