Sicily 8843. Ranking and Friendship

8843. Ranking and Friendship

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

Since teacher Herkabe has started ranking his N students, the number of friendships in his class has sharply fallen. The students near the bottom of the rankings list have become jealous of the top students, while the top students started looking down on their less successful colleagues.
According to Malcolm's observations, the following rule holds: two students are friends if their ranks are close enough, more precisely, if they differ by at most K. For example, if K = 1, then only neighbouring students on the rankings list are friends. Furthermore, two students are good friends if they are friends and their names have the same length.
Write a program to calculate the number of pairs of good friends in this gifted class.

Input

The first line of input contains two positive integers, N (3 ≤ N ≤ 300 000) and K (1 ≤ K ≤ N), from the problem statement.
Each of the following N lines contains a single student's name. The names are given in the order they appear on the rankings list. They consist of between 2 and 20 (inclusive) uppercase English letters.

Output

The first and only line of output must contain the required number of pairs.

Sample Input

样例1：4 2IVAIVOANATOM样例2：6 3CYNTHIALLOYDSTEVIEKEVINMALCOLMDABNEY

Sample Output

样例1：5样例2：2

Problem Source

2013年每周一赛第七场/COCI 2012.12

// Problem#: 8843// Submission#: 3557694// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <algorithm>using namespace std;struct man {    int rank, length;};man M[300005];bool cmp(const man & m1, const man & m2) {    if (m1.length == m2.length) return m1.rank < m2.rank;    return m1.length < m2.length;}inline int ABS(int a) {if (a < 0) return -a; return a;}int main() {    int N, K;    scanf("%d%d\n", &N, &K);    char name[25];    int l = 0;    for (int i = 0; i < N; i++) {        M[i].rank = i;        gets(name);        l = 0;        while (name[l]) l++;        M[i].length = l;    }    sort(M, M + N, cmp);    int startPos = 0, endPos = 0;    long long ans = 0;    for (int i = 1; i < N; i++) {        if (M[startPos].length == M[i].length)            if (ABS(M[startPos].rank - M[i].rank) <= K) {                ans += endPos - startPos + 1;                endPos++;            } else {                while (startPos <= endPos && ABS(M[startPos].rank - M[i].rank) > K) startPos++;                ans += endPos - startPos + 1;                endPos++;            }        else startPos = endPos = i;        }    printf("%lld\n", ans);    return 0;}