Codeforces Round #354 (Div. 2) C. Vasya and String

题目链接：

http://codeforces.com/contest/676/problem/C

题解：

把连续的一段压缩成一个数，对新的数组求前缀和，用两个指针从左到右线性扫一遍。

一段值改变一部分的情况考虑的不够周到，wa了两次。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;

const int maxn = 100000 + 10;
char str[maxn];
int arr[maxn], tot;
int sum[maxn];

int n, k;

int solve(int fst) {
    int ret = -1, cnt = 0;
    int p1 = fst;
    for (int p2 = fst; p2 <= tot; p2 += 2) {
        cnt += arr[p2];
        while (p1 <= p2&&cnt > k) {
            cnt -= arr[p1]; p1 += 2;
        }
        if (p1 <= p2) {
            if (p1 - 2 >= 0 && p2 + 1 <= n) {
                ret = max(ret, sum[p2 + 1] - sum[p1 - 2]+k-cnt);
                ret = min(ret, n);
            }
            else if (p1 - 2 >= 0) {
                ret = max(ret, sum[p2] - sum[p1 - 2] + k - cnt);
                ret = min(ret, n);
            }
            else if (p2 + 1 <= n) {
                ret = max(ret, sum[p2 + 1] - sum[p1 - 1] + k - cnt);
                ret = min(ret, n);
            }
            else {
                ret = max(ret, sum[p2] - sum[p1 - 1] + k - cnt);
                ret = min(ret, n);
            }
        }
    }
    return ret;
}

int main() {
    while (scanf("%d%d", &n, &k) == 2 && n) {
        scanf("%s", str);
        tot = 0;
        int cnt = 1;
        for (int i = 0; i < n - 1; i++) {
            if (str[i] != str[i + 1]) {
                arr[++tot] = cnt;
                cnt = 1;
            }
            else {
                cnt++;
            }
        }
        arr[++tot] = cnt;
        sum[0] = 0;
        for (int i = 1; i <= tot; i++) sum[i] = sum[i - 1] + arr[i];
        int ans = 0;
        for (int i = 1; i <= tot; i++) {
            ans = max(ans, arr[i] + k);
            ans = min(ans, n);
        }
        ans = max(ans, solve(1));
        ans = max(ans, solve(2));
        printf("%d
", ans);
    }
    return 0;
}

/*
4 2
abba
8 1
aabaabaa
5 2
ababa
5 1
ababa
4 1
aaba
5 0
aaaaa
*/