HDU-5672-String（模拟/追赶）

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 918    Accepted Submission(s): 295

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 integerT(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 leastk dictinct characters.

 

Sample Input

2abcabcabca4abcabcabcabc3

 

Sample Output

055

 

Source

BestCoder Round #81 (div.2)

 

String

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))。

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <set>#include <map>#include <queue>#include <stack>using namespace std;#define MAXN 1000005char str[MAXN];long long num[30];int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%s",str+1);        long long len = strlen(str+1);        long long n;        scanf("%I64d",&n);        long long l=1,r=0,tot=0,ans=0;        for(int i=0; i<26; ++i)            num[i]=0;        while(1)        {            if(tot<n && r<len)            {                int x = str[++r]-'a';                if(num[x]==0)++tot;                num[x]++;            }            else if(n<=tot)            {                ans+=len-r+1;                int x = str[l++]-'a';                num[x]--;                if(num[x]==0)--tot;            }            else break;        }        printf("%I64d\n",ans);    }    return 0;}