4553: [Tjoi2016&Heoi2016]序列

4553: [Tjoi2016&Heoi2016]序列

链接

分析：

　　注意所有m此操作中，只会发生一个，于是考虑dp。dp[i]=dp[j]+1，j<i，a[j]<=L[i]，R[j]<=a[i]。L[i]为位置i处，所有可能发生的改变中的最小值，R[i]为最大值。

　　这是三维偏序问题，于是CDQ+树状数组。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005;
int dp[N];
struct Node { int id, x, l, r; } A[N], B[N];
bool cmp1(Node A, Node B) { return A.l < B.l; }
bool cmp2(Node A, Node B) { return A.x < B.x; }
bool cmp3(Node A, Node B) { return A.id < B.id; }

struct Bit{
    int mx[N], n;
    void update(int p,int v) {
        for (; p <= n; p += (p & (-p))) mx[p] = max(mx[p], v);
    }
    int query(int p) {
        int ans = 0;
        for (; p; p -= (p & (-p))) ans = max(ans, mx[p]);
        return ans;
    }
    void clear(int p) {
        for (; mx[p] && p <= n; p += (p & (-p))) mx[p] = 0;
    }
}bit;
void cdq(int l,int r) {
    if (l >= r) return ;
    int mid = (l + r) >> 1;
    cdq(l, mid);
    sort(A + mid + 1, A + r + 1, cmp1);
    int i = l, j = mid + 1;
    while (i <= mid && j <= r) {
        if (A[i].x <= A[j].l) {
            bit.update(A[i].r, dp[A[i].id]); i ++;
        }
        else {
            dp[A[j].id] = max(dp[A[j].id], bit.query(A[j].x) + 1); j ++;
        }
    }
    while (j <= r) {
        dp[A[j].id] = max(dp[A[j].id], bit.query(A[j].x) + 1); j ++;
    }
    for (int k = l; k < i; ++k) bit.clear(A[k].r);
    sort(A + mid + 1, A + r + 1, cmp3);
    cdq(mid + 1, r);
    sort(A + l, A + r + 1, cmp2);
}
int main() {
    int n = read(), m = read(); bit.n = n;
    for (int i = 1; i <= n; ++i) A[i].id = i, A[i].x = A[i].l = A[i].r = read();
    for (int i = 1; i <= m; ++i) {
        int x = read(), y = read();
        A[x].l = min(A[x].l, y), A[x].r = max(A[x].r, y);
    }
    for (int i = 1; i <= n; ++i) dp[i] = 1;
    cdq(1, n);
    int ans = 0;
    for (int i = 1; i <= n; ++i) ans = max(ans, dp[i]);
    cout << ans;
    return 0;
}