hust校赛 f题 The tree of hust（lis 变形）

题目大意是给出一段数字序列，可以忽略一次一段连续的序列，求忽略后的最长连续上升子序列

思路是dp，用end数组记录以当前元素作为结尾的最长连续上升序列的元素个数，那么不难得到状态转移方程为

dp(i) = max(dp(i - 1),  max( end[k] ) ) + 1

代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
using namespace std;
#define LL long long
const int maxn = 200009;
const int INF = 1000000007;

int a[maxn], enda[maxn], d[maxn], g[maxn]; //enda数组记录以当前元素结尾的最长上升连续序列的长度 

int main() {
	int startt = clock();
	freopen("input.txt", "r", stdin);
	int t, kase = 0;
	scanf("%d", &t);
	while(t--) {
		int n; scanf("%d", &n);
		a[0] = 0; enda[0] = 0;
		for(int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
			if(a[i] > a[i - 1]) enda[i] = enda[i - 1] + 1;
			else enda[i] = 1;
		//	printf("%d\n", enda[i]);
		}

		int ans = 0;
		for(int i = 1; i <= n; i++) g[i] = INF;  //lis的变形，g[i]表示enda的值为i的最小元素值
		for(int i = 1; i <= n; i++) {
			int k = lower_bound(g + 1, g + n + 1, a[i]) - g - 1;

			if(a[i] > a[i - 1])	d[i] = max(d[i - 1], k) + 1;
			else d[i] = k + 1;

			ans = max(ans, d[i]);
			g[enda[i]] = min(g[enda[i]], a[i]);
		//	printf("%d\n", d[i]);
		} 

		printf("Case #%d: %d\n", ++kase, ans);
	}
	int endt = clock();
	printf("%d\n", endt - startt);
	return 0;
}