HDU 4339 Query (线段树)

Query

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1922    Accepted Submission(s): 654

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

 

Source

2012 Multi-University Training Contest 4

 

Recommend

zhoujiaqi2010

题意： 有两个字符串，给出 Q 个询问，每个询问有两种体式格式：

1 p i c 把第 p 个字符串的第i 个字符换成 字符 c，

2 i     从第位i 开始，两个字符串连续相同的子串的最大长度为多少。

思路：普通的查找肯定超时，所以用线段树 复杂度大概为O（n + Qlog(n)） 每个结点保存该区间内第一个不匹配的下标，query时，只需找到与key（要查找的点） 最近的一个下标 相减就得

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

using namespace std;

#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)

const int maxn=1000010;

struct node{
    int l,r;
    int lab;    //lab表示在当前区间内第一个不匹配的下标
}tree[maxn*3];

char str[2][maxn];
int tmp,a,b;

void build(int l,int r,int rt){
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].lab=tmp;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(l,mid,L(rt));
    build(mid+1,r,R(rt));
}

void update(int key,int lab,int rt){
    if(tree[rt].l==tree[rt].r){
        tree[rt].lab=lab;
        return ;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(key<=mid)    
        update(key,lab,L(rt));
    else
        update(key,lab,R(rt));
    tree[rt].lab=min(tree[L(rt)].lab,tree[R(rt)].lab);
}

int query(int key,int rt){
    if(key<=tree[rt].l) //只需要找>= key 的范围
        return tree[rt].lab;
    if(key<=tree[L(rt)].r)
        a=query(key,L(rt)); //a 原初始化为tmp
    b=query(key,R(rt)); //右边的肯定满足
    return min(a,b);    
}

int main(){

    //freopen("input.txt","r",stdin);

    int T,Q;
    char c;
    scanf("%d",&T);
    int cases=0;
    while(T--){
        scanf("%s%s",str[0],str[1]);
        int len=min(strlen(str[0]),strlen(str[1]));
        tmp=len+1;
        build(1,len,1);
        for(int i=0;i<len;i++)
            if(str[0][i]!=str[1][i])
                update(i+1,i+1,1);
            else
                update(i+1,tmp,1);
        scanf("%d",&Q);
        printf ("Case %d:\n",++cases);
        int op,s,k;
        while(Q--){
            scanf("%d",&op);
            if(op==2){
                scanf("%d",&k);
                a=tmp;
                int ans=query(k+1,1)-(k+1);
                printf("%d\n",ans);
            }else{
                scanf("%d %d %c",&s,&k,&c);
                s--;    //只有两种情况需要更新
                if(str[s][k]==str[s^1][k] && str[s^1][k]!=c)    //原来两字符相等 改后不等了
                    update(k+1,k+1,1);
                if(str[s][k]!=str[s^1][k] && str[s^1][k]==c)    //原两字符不等 改后相等了
                    update(k+1,tmp,1);
                str[s][k]=c;
            }
        }
    }
    return 0;
}