795. 区间子数组个数

Posted 2023-04-05 magic_zk

题目描述

给一个数组，再给一个值的范围[l, r]，
问最大值在[l, r]之间的子数组有多少个？

f1-双指针

基本分析

1. 如果枚举子数组的右端点i，会有几种情况？（1）arr[i] > right; (left <= arr[i] <= right; (3)arr[i] < left
2. 假如枚举到右端点i，左端点怎么考虑？（1）的情况，这个子数组不满足，可以跳过；(2)的情况i可以是一个最近的左端点，只需要找到最远的左端点i就行。分两种情况，前面没有>right的索引，i就是0；前面有>right的索引，i就是索引+1；（3）的情况，自己不能当左端点，除了找最远的左端点以外，还要找最近的左端点i’，可选的左端点个数是i\' - i
3. 怎么实现以上逻辑？用两个指针last1表示最近的左端点位置；last2表示最近的>right的位置。对枚举到的每个值，当last1存在的时候，表示有左端点存在，就可以累加结果

代码

class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
        last1 = last2 = -1
        ans = 0

        for i, x in enumerate(nums):
            if left <= x <= right:
                last1 = i
            elif x > right:
                last2 = i
                last1 = -1
            if last1 != -1:
                ans += last1 - last2
        
        return ans

总结

1. 枚举右端点，右端点可以保证子数组不同
2. 个数取决于可选的左端点范围，这里用last1和last2来维护

f2-计数

基本分析

1. 怎么用计数的角度考虑？（1）找<= right的区间个数（2）找<=left-1的区间个数；（3）求差
2. 怎么统计<=x的区间个数？如果长度是l，个数就是1+2+3+...+l,碰到不满足的重新计数。

代码

class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:

        def cal(x):
            ans = 0
            cnt = 0
            for num in nums:
                if num <= x:
                    cnt += 1
                else:
                    cnt = 0
                ans += cnt
            return ans
        
        return cal(right) - cal(left - 1)

总结

1. 利用题目的要求，压缩每个数字的含义，使其变为简单的0、1和2
2. 利用容斥的思想，使用计数解决

f3-单调栈

待补充

基本分析

1. xxx

代码

总结

1. xxx

HDU4622:Reincarnation(后缀数组，求区间内不同子串的个数)

Problem Description

Now you are back,and have a task to do:
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.

 

Input

The first line contains integer T(1<=T<=5), denote the number of the test cases.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.

 

Output

For each test cases,for each query,print the answer in one line.

 

Sample Input

2
bbaba
5
3 4
2 2
2 5
2 4
1 4
baaba
5
3 3
3 4
1 4
3 5
5 5

 

Sample Output

3
1
7
5
8
1
3
8
5
1
Hint
      I won‘t do anything against hash because I am nice.Of course this problem has a solution that don‘t rely on hash.
 

 

Author

WJMZBMR

 

Source

2013 Multi-University Training Contest 3

 

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题意：

求区间内不同子串的个数

思路：

论文里面有求整个串的不同子串的个数。我们能够引申到这道题

对于整个串，我们的求法是全部子串数减去全部height的值，而height就是lcp

那么对于某个区间，我们仅仅要求出全部包括在这个区间的后缀。然后减去互相之间的lcp就可以

关键是我们要保持这个区间的后缀的字典序

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 2005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank[N],height[N],s[N],a[N];
char str[N],str1[N],str2[N];
#define F(x) ((x)/3+((x)%3==1?
0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
int cmp(int *r,int a,int b,int k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包括末尾加入的0
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++)  wsf[i]=0;
    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
    p=1;
    j=1;
    for(; p<n; j*=2,m=p)
    {
        for(p=0,i=n-j; i<n; i++)  y[p++]=i;
        for(i=0; i<n; i++)  if(sa[i]>=j)  y[p++]=sa[i]-j;
        for(i=0; i<n; i++)  wv[i]=x[y[i]];
        for(i=0; i<m; i++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[wv[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
        t=x;
        x=y;
        y=t;
        x[sa[0]]=0;
        for(p=1,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
    }
}
void getheight(int *r,int n)//n不保存最后的0
{
    int i,j,k=0;
    for(i=1; i<=n; i++)  rank[sa[i]]=i;
    for(i=0; i<n; i++)
    {
        if(k)
            k--;
        else
            k=0;
        j=sa[rank[i]-1];
        while(r[i+k]==r[j+k])
            k++;
        height[rank[i]]=k;
    }
}

int Log[N];
int best[30][N];

void setLog()
{
    Log[0] = -1;
    for(int i=1; i<N; i++)
    {
        Log[i]=(i&(i-1))?Log[i-1]:Log[i-1] + 1 ;
    }
}
void RMQ(int n)  //初始化RMQ
{
    for(int i = 1; i <= n ; i ++) best[0][i] = height[i];
    for(int i = 1; i <= Log[n] ; i ++)
    {
        int limit = n - (1<<i) + 1;
        for(int j = 1; j <= limit ; j ++)
        {
            best[i][j] = min(best[i-1][j] , best[i-1][j+(1<<i>>1)]);
        }
    }
}
int LCP(int a,int b)  //询问a,b后缀的最长公共前缀
{
    a ++;
    int t = Log[b - a + 1];
    return min(best[t][a] , best[t][b - (1<<t) + 1]);
}
int t,n,m;

int solve(int l,int r,int n)
{
    int ans = (r-l+1)*(r-l+2)/2;
    int last = -1;
    int cnt = r-l+1;
    for(int i = 1; i<=n; i++)
    {
        if(!cnt) break;
        if(sa[i]<l || sa[i]>r) continue;
        cnt--;
        if(last == -1)
        {
            last = i;
            continue;
        }
        int a = last;
        int b = i;
        if(a>b) swap(a,b);
        int lcp = LCP(a,b);
        int la = r-sa[last]+1;//区间内该串的尾部
        int lb = r-sa[i]+1;
        if(la>=lb && lcp>=lb);//la包括lb了，那么就用la继续往后比較，借此保持字典序，来模拟得到该区间的全部height
        else last = i;
        ans-=min(lcp,min(la,lb));
    }
    return ans;
}

int main()
{
    int i,j,k,len,l,r;
    scanf("%d",&t);
    setLog();
    W(t--)
    {
        scanf("%s",str);
        scanf("%d",&m);
        len = strlen(str);
        for(i = 0; i<len; i++)
            s[i] = str[i]-'a'+1;
        s[len] = 0;
        getsa(s,sa,len+1,30);
        getheight(s,len);
        RMQ(len);
        while(m--)
        {
            scanf("%d%d",&l,&r);
            printf("%d\n",solve(l-1,r-1,len));
        }
    }
    return 0;
}