hdu4339 Query（multiset 或 线段树）

Query

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 998    Accepted Submission(s): 300

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

 

题意：第一行输入一个整数，表示案例数；然后输入两个字符串；再然后输入一个数，表示操作的次数；

两个操作：

1 a i c  更新：表示第 a（a=1或a=2）个字符串中下标为 i 的字符用 c 替换；

2 i       访问：表示两个字符串从下标为 i 开始有多个字符是一样的。

 

法一：把所以不能匹配的下标存储在 multiset

#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#define MAXN 1000010

using namespace std;

char s[2][MAXN];

int main()
{
    int test,n,i,j,len1,len2,p,q,k=1;
    char c[2];
    multiset<int>ms;
    multiset<int>::iterator it;
    scanf("%d",&test);
    while(test--)
    {
        ms.clear();
        scanf("%s",s[0]);
        scanf("%s",s[1]);
        len1=strlen(s[0]);
        len2=strlen(s[1]);
        i=0;
        while(i<len1&&i<len2)
        {
            if(s[0][i]!=s[1][i]) ms.insert(i);
            i++;
        }
        ms.insert(i);
        printf("Case %d:\n",k++);
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&p);
            if(p==2)
            {
                scanf("%d",&j);
                it=ms.lower_bound(j);
                printf("%d\n",*it-j);
            }
            else
            {
                scanf("%d %d %s",&q,&j,c);
                q--;
                if(s[0][j]!=s[1][j]) ms.erase(j);
                s[q][j]=*c;
                if(s[q^1][j]!=s[q][j]) ms.insert(j);
            }
        }
    }
    return 0;
}

 

法二：线段树中存储这区间第一个不能匹配的下标，转换成单点更新，区间最值

#include<iostream>
#include<cstring>
#include<cstdio>
#define MAXN 1000000

struct segtree
{
    int left,right,acount;
}tree[MAXN<<2];
char s[2][MAXN];
int INF;

int min(int A,int B){return A<B?A:B;}

int max(int A,int B){return A>B?A:B;}

void build(int r,int a,int b)
{
    tree[r].left=a;tree[r].right=b;
    tree[r].acount=INF;
    if(a<b)
    {
        int m=(a+b)>>1;
        build(r<<1,a,m);
        build(r<<1|1,m+1,b);
    }
}

void update(int r,int i,int key)
{
    if(tree[r].left==tree[r].right)
    {
        tree[r].acount=key;
        return ;
    }
    int m=(tree[r].left+tree[r].right)>>1;
    if(i<=m) 
        update(r<<1,i,key);
    else 
        update(r<<1|1,i,key);
    tree[r].acount=min(tree[r<<1].acount,tree[r<<1|1].acount);
}

int find_min(int r,int a,int b)
{
    if(a<=tree[r].left&&tree[r].right<=b)
        return tree[r].acount;  
    int m=(tree[r].left+tree[r].right)>>1;
    if(b<=m) 
        return find_min(r<<1,a,b);//对自己很无语了居然写成find_min(r<<1,a,m);
    else if(a>m) 
        return find_min(r<<1|1,a,b);//这里也是
    else 
        return min(find_min(r<<1,a,m),find_min(r<<1|1,m+1,b));
}

int main()
{

#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif

    int test,i,j,k=1,n,p,q,len;
    char c[2];
    scanf("%d",&test);
    while(test--)
    {
        scanf("%s",s[0]);
        scanf("%s",s[1]);
        len=min(strlen(s[0]),strlen(s[1]));
        INF=len+1;
        build(1,1,len);
        for(i=0;i<len;i++)
        {
            if(s[0][i]!=s[1][i]) update(1,i+1,i+1);
            else update(1,i+1,INF);
        }
        printf("Case %d:\n",k++);
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&p);
            if(p==2)
            {
                scanf("%d",&j);
                if(j>=len)
                    printf("0\n");
                else
                    printf("%d\n",find_min(1,j+1,len)-j-1);
            }
            else if(p==1)
            {
                scanf("%d %d %s",&q,&j,c);
                if(j>=len) continue;
                q--;
                if(s[q][j]==s[q^1][j]&&s[q^1][j]!=*c) update(1,j+1,j+1);
                if(s[q][j]!=s[q^1][j]&&s[q^1][j]==*c) update(1,j+1,INF);
                s[q][j]=*c;
            }
        }
    }
    return 0;
}