HDOJ 题目4339 Query（线段树，单点更新）

最新推荐文章于 2015-11-15 17:23:25 发布

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本文介绍了一种处理字符串匹配及动态更新的算法实现。通过构建特殊的数据结构来高效地完成两个字符串之间的匹配，并能实时更新字符串内容以反映最新的匹配状态。此算法适用于字符串长度较大且查询次数频繁的应用场景。

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Query

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2859 Accepted Submission(s): 925

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

Source

2012 Multi-University Training Contest 4

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题目大意：两种操作，1 2，操作为1时输入 a b c ,第a个字符串的第b位变成c，操作为2时，输入a ，询问从第a位起两个字符串有多少位是相等的

ac代码

#include<stdio.h>
#include<string.h>
#define min(a,b) (a>b?b:a)
int node[1000010<<2];
char s1[1000010],s2[1000010];
void pushup(int tr)
{
	node[tr]=node[tr<<1]+node[tr<<1|1];
}
void build_tr(int l,int r,int tr)
{
	if(l==r)
	{
		if(s1[l]==s2[l])
		{
			node[tr]=1;
		}
		else
			node[tr]=0;
		return;
	}
	int mid=(l+r)>>1;
	build_tr(l,mid,tr<<1);
	build_tr(mid+1,r,tr<<1|1);
	pushup(tr);
}
void update(int pos,int l,int r,int tr)
{
	if(l==r)
	{
		if(s1[l]==s2[l])
		{
			node[tr]=1;
		}
		else
			node[tr]=0;
		return;
	}
	int mid=(l+r)>>1;
	if(pos<=mid)
	{
		update(pos,l,mid,tr<<1);
	}
	else
		update(pos,mid+1,r,tr<<1|1);
	pushup(tr);
}
int query(int pos,int l,int r,int tr)
{
	if(l==r)
		return node[tr];
	if(pos>=l&&pos<=r)
	{
		if(node[tr]==r-l+1)
		{
			return r-pos+1;
		}
		else
		{
			int mid=(l+r)>>1;
			int ans=0;
			if(pos<=mid)
			{
				ans+=query(pos,l,mid,tr<<1);
				if(ans==mid-pos+1)
					ans+=query(mid+1,mid+1,r,tr<<1|1);
			}
			else
				ans+=query(pos,mid+1,r,tr<<1|1);
			return ans;
		}
	}
}
int main()
{
	int t,c=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%s",s1+1,s2+1);
		int len1=strlen(s1+1);
		int len2=strlen(s2+1);
		int len=min(len1,len2);
		build_tr(1,len,1);
		int q;
		scanf("%d",&q);
		printf("Case %d:\n",++c);
		while(q--)
		{
			int op;
			scanf("%d",&op);
			if(op==1)
			{
				char s[2];
				int a,b;
				scanf("%d%d%s",&a,&b,s);
				if(a==1)
					s1[b+1]=s[0];
				else
					s2[b+1]=s[0];
				update(b+1,1,len,1);
			}
			else
			{
				int a;
				scanf("%d",&a);
				printf("%d\n",query(a+1,1,len,1));
			}

		}
	}
}