BestCoder Round #81 (div.2)C HDOJ5672 String(双指针法)

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本文介绍了一种算法，用于解决给定字符串中包含至少k个不同字符的子串数量计算问题。通过移动指针技术和哈希映射，有效地解决了这一挑战。

String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 813 Accepted Submission(s): 258

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

题目链接:点击打开链接

给出一个字符串和需要的字符串长度, 输出复合要求子串个数.

头尾指针移动, 等于需要字符串长度时计算尾指针距离末尾长度, 得到子串个数累加.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define st first
#define nd second
#define exp 1e-8
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e6 + 6;
int t, k, len, st, ed, cnt, num[30];
char s[MAXN];
std::map<char, int> mp;
int main(int argc, char const *argv[])
{
    scanf("%d", &t);
    while(t--) {
        memset(num, 0, sizeof(num));
        scanf("%s%d", s, &k);
        len = strlen(s), st = 0, ed = 0, cnt = 0;
        ll ans = 0;
        while(st < len) {
            if(st) {
                num[s[st - 1] - 'a']--;
                if(!num[s[st - 1] - 'a']) cnt--;
            }
            while(cnt < k && ed < len) {
                if(!num[s[ed] - 'a']) cnt++;
                num[s[ed] - 'a']++;
                ed++;
            }
            if(cnt >= k) ans += len - ed + 1;
            st++;
        }
        printf("%lld\n", ans);
    }
    return 0;
}