nowcoder911L 最优子区间

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;
}