HDU 5672 String（尺取法）

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概述

Problem Description

There is a string

S .

S only contain lower case English character.

(10≤length(S)≤1,000,000)
How many substrings there are that contain at least

k(1≤k≤26) distinct characters?

Input

There are multiple test cases. The first line of input contains an integer

T(1≤T≤10) indicating the number of test cases. For each test case:

The first line contains string

S .
The second line contains a integer

k(1≤k≤26) .

Output

For each test case, output the number of substrings that contain at least

k dictinct characters.

Sample Input

  
  
   
   2
abcabcabca
4
abcabcabcabc
3

Sample Output

Source

BestCoder Round #81 (div.2)

题意:给你一个字符串，问你有多少子串包含k个不同的字母。

思路：如果有一段字符刚刚好满足条件，那么后面的包含这个串的子串全部满足，我们可以尺取l和r，对于一个r满足的话，后面len - r + 1个子串也满足，然后更新l就可以了，因为l和r是分开更新的，所以复杂度是O(n)。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const double  pi = acos(-1.0);
const int N = 1e6 + 10;
char s[N];
int main()
{
    int t, k, vis[40];
    cin>>t;
    while(t--)
    {
        scanf("%s", s);
        scanf("%d", &k);
        memset(vis, 0, sizeof(vis));
        int l = 0, r = 0, len = strlen(s), temp, num = 0;
        ll ans = 0;
        while(l<len)
        {
            while(r < len && num < k)
            {
                temp = s[r] - 'a';
                if(!vis[temp])
                    num++;
                vis[temp]++;
               r++;
            }
            if(num<k)
                break;
            ans += len - r + 1;
            temp = s[l] - 'a';
            vis[temp]--;
            if(vis[temp]==0)
                num--;
            l++;
        }
        cout<<ans<<endl;
    }
    return 0;
}