nowcoder911L 最优子区间

题目链接

思路

用\(f(i,j)\)表示前i个元素，以i为右端点，j为左端点时的答案。

用个"区间修改,单点查询"的线段树维护出第二维。在从左往右枚举i的过程中。将\([lst_i+1,i]\)的答案+1.将\([lst_{lst_i}+1,lst_i]\)的答案-1。

代码

/*
* @Author: wxyww
* @Date:   2019-06-05 11:13:19
* @Last Modified time: 2019-06-05 11:39:04
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 100000 + 100;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int tree[N << 2],lazy[N << 2];
void pushdown(int rt) {
	if(lazy[rt]) {
		tree[rt << 1] += lazy[rt];
		tree[rt << 1 | 1] += lazy[rt];
		lazy[rt << 1] += lazy[rt];
		lazy[rt << 1 | 1] += lazy[rt];
		lazy[rt] = 0;
	}
}
void update(int rt,int l,int r,int L,int R,int c) {
	if(l >= L && r <= R) {
		lazy[rt] += c;
		tree[rt] += c;
		return;
	}
	pushdown(rt);
	int mid = (l + r) >> 1;
	if(L <= mid) update(rt << 1,l,mid,L,R,c);
	if(R > mid) update(rt << 1 | 1,mid + 1,r,L,R,c);
	tree[rt] = max(tree[rt << 1],tree[rt << 1 | 1]);
}
int pos[N],ans,lst[N];
int main() {
	int n = read();
	int ans = 0;
	for(int i = 1;i <= n;++i) {
		int x = read();
		if(pos[x]) lst[i] = pos[x];
		pos[x] = i;
		if(lst[i]) update(1,1,n,lst[lst[i]] + 1,lst[i],-1);
		update(1,1,n,lst[i] + 1,i,1);
		ans = max(ans,tree[1]);
	}
	cout<<ans;
	return 0;
}