HDU 5672 String（尺取法）

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

 

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
;
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;
}