HDU -5672 String（尺取法）_hdu5672卡常-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_44606097/article/details/99603745

本文介绍了一种算法，用于解决在给定字符串中查找包含至少k个不同字符的所有子串的问题。通过遍历字符串并使用滑动窗口技术，该算法有效地计算了满足条件的子串总数。

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String

Time Limit: 2000/1000 MS (Java/Others)	Memory Limit: 65536/65536 K (Java/Others)

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
0
55

从前往后遍历，统计新出现的字母，若达到k个，则后面的再往上加都满足(ans+=l-r)，然后去掉前i个字符看还有没有k个不同的字符，若有，加上相应的长度，若没有，继续往后遍历

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

char s[1000200];
int vis[30]={0};

int main()
{
    int t, k;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%s%d", s, &k);
        int sum=0, len=strlen(s), r=0;
        long long ans=0;
        for(int i=0; i<len; i++)
        {
            while(sum<k&&r<len)
            {
                vis[s[r]-'a']++;
                if(vis[s[r]-'a']==1)//统计不同的字符
                    sum++;
                r++;
            }
            if(sum==k)//加上后面的长度
                ans+=len-r+1;
            vis[s[i]-'a']--;//去掉前i个字符
            if(vis[s[i]-'a']==0)//若这个字符是唯一的，sum--
                sum--;
        }
        printf("%lld\n", ans);
    }
    return 0;
}