bzoj4282 慎二的随机数列 树状数组求LIS + 构造

首先，我们不难发现N个位置都选一定不会比少选任意几个差，所以我们就先设定我们将这N个修改机会都用上， 那么如果点 i">ii 前有sumv">sumvsumv个可修改点要被选的话，当前点被选择的条件是减掉sumv">sumvsumv后依然能和前面已减掉过sumv">sumvsumv的进行匹配。

Code：

#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 4;
int p, maxv[maxn], tags[maxn], nums[maxn], val[maxn];
inline int lowbit(int t)
{
    return t & (-t);
}
inline void update(int x,int delta)
{
    while(x <= p + 2)
        maxv[x] = max(maxv[x], delta), x += lowbit(x);
}
inline int query(int x)
{
    int ans = 0;
    while(x > 0) ans = max(ans, maxv[x]) , x -= lowbit(x);
    return ans;
}
int main()
{
    freopen("calligraphy9.in","r",stdin);
    //freopen("calligraphy.","w",stdout);
    int n, cnt = 0, sumv = 0, ans = 0, fin = 0;
    scanf("%d",&n);
    for(int i = 1;i <= n; ++i)
    {
        char g[10];
        scanf("%s",g);
        if(g[0] == 'N')
        {
            tags[i] = 1;
            ++sumv;
            continue;
        }
        scanf("%d",&val[i]);
    }
    ans = sumv;
    for(int i = n;i >= 1;--i)
    {
        if(tags[i]) --sumv;
        else val[i] -= sumv;
    }
    for(int i = 1;i <= n; ++i)if(!tags[i]) nums[++cnt] = val[i];
    sort(nums + 1, nums + 1 + cnt);
    for(int i = 1;i <= n; ++i)
        if(!tags[i])
        {
            val[i] = lower_bound(nums + 1, nums + 1 + cnt, val[i]) - nums;
            p = max(p, val[i]);
        }
    for(int i = 1;i <= n; ++i)
    {
        if(tags[i]) continue;
        int u = val[i];
        int pre = query(u - 1);
        fin = max(fin, pre + 1);
        update(u, pre + 1);
    }
    printf("%d",ans + fin);
    fclose(stdin);
    return 0;
}