Codeforces 814C-An impassioned circulation of affection
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Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!
Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from left to right, and the i-th piece has a colour si, denoted by a lowercase English letter. Nadeko will repaint at most m of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c — Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.
For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o". Among all subsegments containing pieces of "o" only, "ooo" is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.
But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has q plans on this, each of which can be expressed as a pair of an integer mi and a lowercase letter ci, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.
The first line of input contains a positive integer n (1 ≤ n ≤ 1 500) — the length of the garland.
The second line contains n lowercase English letters s1s2... sn as a string — the initial colours of paper pieces on the garland.
The third line contains a positive integer q (1 ≤ q ≤ 200 000) — the number of plans Nadeko has.
The next q lines describe one plan each: the i-th among them contains an integer mi (1 ≤ mi ≤ n) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci — Koyomi's possible favourite colour.
Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.
6koyomi31 o4 o4 m
365
15yamatonadeshiko101 a2 a3 a4 a5 a1 b2 b3 b4 b5 b
3457812345
10aaaaaaaaaa210 b10 z
1010
In the first sample, there are three plans:
- In the first plan, at most 1 piece can be repainted. Repainting the "y" piece to become "o" results in "kooomi", whose Koyomity of 3is the best achievable;
- In the second plan, at most 4 pieces can be repainted, and "oooooo" results in a Koyomity of 6;
- In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in a Koyomity of 5.
题意:给你一个字符串,有q次询问,每次询问可以改变m个字符,字符c最多有多少个连续的
解题思路:先预处理出每种字符出现的位置有哪些,每次询问时可以暴力枚举这种字符出现的位置,然后二分向右最远可以覆盖到的同一个字符,多余的字符可以向左向右拓展,若放得下,更新一下答案,因为字符串最长为1500,询问最多有26*1500种可能,远小于题目最多的询问次数,所以可以每次将答案保存下来
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int a[30], b[30][1500], n, m, ans[30][1550];char ch[1550], s[5];int check(int k, int xx, int x, int &sum){sum = b[k][x] - b[k][xx] + 1 - (x - xx + 1);if (sum <= m) return 1;else return 0;}int main(){while (~scanf("%d", &n)){scanf("%s", ch);memset(ans, -1, sizeof ans);memset(a, 0, sizeof a);for (int i = 0; i < n; i++){int k = ch[i] - 'a';b[k][a[k]++] = i;}int q;scanf("%d", &q);while (q--){scanf("%d%s", &m, s);int k = s[0] - 'a', ma = m;if (ans[k][m] != -1) { printf("%d\n", ans[k][m]); continue; }for (int i = 0; i < a[k]; i++){int l = i, r = a[k] - 1;while (l <= r){int mid = (l + r) >> 1, sum = 0;if (check(k, i, mid, sum)){l = mid + 1;if (m - sum <= n - b[k][mid] - 1) ma = max(ma, b[k][mid] - b[k][i] + 1 + m - sum);else{int xx = m - sum - (n - b[k][mid] - 1);ma = max(ma, b[k][mid] - b[k][i] + 1 + n - b[k][mid] - 1 + min(xx, b[k][i]));}}else r = mid - 1;}}ans[k][m] = ma;printf("%d\n", ma);}}return 0;}
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