HDU 5672 String

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1977    Accepted Submission(s): 638

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

2abcabcabca4abcabcabcabc3

 

Sample Output

055

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5672

题意：给定一个字符串和一个k，统计有多少个连续区间内不同字符数大于等于k。

解题思路：尺取法，如果下标s——t（t为第一个从s出发满足条件的下标）满足条件，那么t往后的下标都满足，再将左区间往后挪一，继续判断。

AC代码：

#include<cstdio>#include<cstring>#include<map>#include<algorithm>using namespace std;const int LEN = 1000000 + 10;//输入字符串的最长长度 int k;char str[LEN];void solve(){int s,t;int num;//统计查找区间不同字符的个数 long long cnt;//定义为long long型，因为字符串长度有1000000 int n;n=strlen(str);//字符串的长度 s=0,t=0;num=0,cnt=0;map<char,int> count;//声明map用来记录每个字符出现的次数 for(;;)//核心算法 ，尺取法{while(t<n&&num<k)//当下标小于字符穿长度，且区间中不同字符数小于k时循环 {if(count[str[t++]]++==0)//出现新字符 {num++;}}if(num<k) break;//当查找区间内不同字符数小于k也即t到达字符串末尾时跳出循环 cnt+=n+1-t;//更新cnt，下标t满足条件，那么t往后的都满足 if(--count[str[s]]==0) num--;//判断左区间字符出现的次数 s++;//将左区间往后挪一}printf("%lld\n",cnt);}int main(void){int T;//测试组数 scanf("%d",&T);while(T--){getchar();scanf("%s%d",str,&k);solve();}return 0;}