BestCoder Round #81 (div.2)

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 454    Accepted Submission(s): 138

有一个明显的性质：如果子串$(i,j)$包含了至少$k$个不同的字符，那么子串$(i,k),(j<k<length)$也包含了至少$k$个不同字符。

因此对于每一个左边界，只要找到最小的满足条件的右边界，就能在$O(1)$时间内统计完所有以这个左边界开始的符合条件的子串。

寻找这个右边界，是经典的追赶法（尺取法,双指针法）问题。维护两个指针（数组下标），轮流更新左右边界，同时累加答案即可。复杂度 $O(length(S))$。

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)

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>
#include <set>
#define LL long long
using namespace std;
char s[1000010];
int A[128];
int cnt;
LL ans;
int main()
{
    int t,len,p;
    cin>>t;
    while(t--)
    {
        scanf("%s",s);
        scanf("%d",&p);
        memset(A,0,sizeof(A));
        ans=0;
        cnt=0;
        len=strlen(s);
        for(int i=0,j=0;i<len;i++)
        {
            while(j<len&&cnt<p)
            {
                A[s[j]]++;
                if(A[s[j]]==1)
                {
                    cnt++;
                }
                j++;
            }
            if(cnt>=p)
            ans+=len-j+1;
            else
                break;
            A[s[i]]--;
            if(A[s[i]]==0)
                cnt--;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}