hdu4339 Query(简单线段树)

Query

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2059    Accepted Submission(s): 706

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
   
   
   
   

题目分析：给2个字符串，2个操作，第一个操作就将某个字符串某个位置的字符换掉。第二个操作是查询2个字符串中某个位置开始向右最长连续相等的长度。

题目分析：线段树。比赛的时候竟然没做出来，还是太水了啊。。。

每个区间维护2个值：lc、rc分别表示当前区间左右两边连续相等的最长长度。查询的时候，假设查询p位置，那么只需要讨论p的位置即可。

对于一个区间，p的位置只有3种可能：

如图，当p在p1位置的时候，直接返回a1，当p在p3位置的时候，显然a2是答案的一部分，但是有可能右边还有解，所以还要再查询当前区间右端点的右边一个点，当p在p2位置的时候，p的位置还是不确定的，所以继续向下查询，缩小p的范围，直到能确定p的位置为止。

详情请见代码：

#include <iostream>
#include<cstdio>
#include<cstring>

using namespace std;
const int N = 1000005;
char s1[N],s2[N];
struct node
{
    int lc,rc;//当前区间左右连续相等
}tree[N<<2];
int l1,l2;
int n,m;

int Min(int a,int b)
{
    return a < b?a:b;
}

void pushup(int num,int s,int e)
{
    int ls = num<<1;
    int rs = num<<1|1;
    int mid = (s + e)>>1;
    tree[num].lc = tree[ls].lc;
    if(tree[ls].lc == mid - s + 1)
        tree[num].lc += tree[rs].lc;
    tree[num].rc = tree[rs].rc;
    if(tree[rs].rc == e - mid)
        tree[num].rc += tree[ls].rc;
}

void build(int num,int s,int e)
{
    tree[num].lc = tree[num].rc = 0;
    if(s == e)
    {
        if(s1[s - 1] == s2[s - 1])
        {
            tree[num].lc = tree[num].rc = 1;
        }
        return;
    }
    int mid = (s + e)>>1;
    build(num<<1,s,mid);
    build(num<<1|1,mid + 1,e);
    pushup(num,s,e);
}

void update(int num,int s,int e,int pos)
{
    if(pos > e)
        return;
    if(s == e)
    {
        if(s1[s - 1] == s2[s - 1])
        {
            tree[num].lc = tree[num].rc = 1;
        }
        else
            tree[num].lc = tree[num].rc = 0;
        return;
    }
    int mid = (s + e)>>1;
    if(pos <= mid)
        update(num<<1,s,mid,pos);
    else
        update(num<<1|1,mid + 1,e,pos);
    pushup(num,s,e);
}

int query(int num,int s,int e,int pos)
{
    if(pos > e)
        return 0;
    if(s1[pos - 1] != s2[pos - 1])
        return 0;
    if(pos - s + 1 <= tree[num].lc)
    {
        return tree[num].lc - pos + s;
    }
    if(e - pos + 1 <= tree[num].rc)
    {
        return e - pos + 1 + query(1,1,n,e + 1);
    }
    int mid = (s + e)>>1;
    if(pos <= mid)
        return query(num<<1,s,mid,pos);
    else
        return query(num<<1|1,mid + 1,e,pos);
}

int main()
{
    int t;
    int op,a,b;
    char ch[3];
    scanf("%d",&t);
    int cas = 0;
    while(t --)
    {
        scanf("%s",s1);
        scanf("%s",s2);
        l1 = strlen(s1);
        l2 = strlen(s2);
        n = Min(l1,l2);
        build(1,1,n);
        scanf("%d",&m);
        printf("Case %d:\n",++cas);
        while(m --)
        {
            scanf("%d",&op);
            if(op == 2)
            {
                scanf("%d",&a);
                printf("%d\n",query(1,1,n,a + 1));
            }
            else
            {
                scanf("%d%d%s",&a,&b,ch);
                if(a == 1)
                    s1[b] = ch[0];
                else
                    s2[b] = ch[0];
                update(1,1,n,b + 1);
            }
        }
    }
    return 0;
}
//1843MS	18624K

特别鸣谢：

特别感谢@男神帮忙找到了一直ole的bug。。。

最近好不顺啊啊。。。