学步园

题目链接：uva 1471 - Defense Lines

题目大意：给定一个序列，删除一段连续的序列后，使得现在序列中连续的递增序列尽量长。

解题思路：维护单调即可，即连续长度最长的情况下结尾的小值，然后对于新的一段，用二分找到满足的连接处。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 2 * 1e5 + 5;
const int INF = 0x3f3f3f3f;

int N, rec[maxn], arr[maxn];

int bsearch (int v) {
    int l = 0, r = N + 1;
    for (int i = 0; i < 100; i++) {
        int mid = (l + r) / 2;
        if (rec[mid] < v)
            l = mid;
        else
            r = mid;
    }
    return l;
}

int solve () {
    memset(rec, INF, sizeof(rec));
    rec[0] = 0;
    int pre = 0, c = 0, ret = 0;
    for (int i = 1; i <= N; i++) {

        if (arr[i] > pre) {
            c++;
        } else {
            do {
                ret = max(ret, c + bsearch(arr[i-c]));
            } while (--c);

            c = 1;
        }
        rec[c] = min(rec[c], arr[i]);
        pre = arr[i];
    }

    while (c--)
        ret = max(ret, c + bsearch(arr[N-c]) + 1);
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d", &N);
        for (int i = 1; i <= N; i++)
            scanf("%d", &arr[i]);
        printf("%d\n", solve());
    }
    return 0;
}