SPOJ Another Longest Increasing Subsequence Problem

传送门：https://www.spoj.com/problems/LIS2/en/

题意：

给定 N个数对 \((x_i,y_i)\)，求最长上升子序列的长度。上升序列定义为满足\((x_i,y_i)\)对i<j 有 \(x_i<x_j\) 且 \(y_i<y_j\)

题解：

一个三维最长链问题

第一维是存位置，第二维存x，第三维存y

注意查询是查询到p[i].z-1然后更新

细节方面和HDU4742是一样的

详情：https://www.cnblogs.com/buerdepepeqi/p/11193426.html

代码：

#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
struct node {
    int x, y, z;
} p[maxn];
bool cmpx(node A, node B) {
    return A.x < B.x;
}
bool cmpy(node A, node B) {
    return A.y < B.y;
}
int lowbit(int x) {
    return x & (-x);
}
int n;
int dp[maxn];
int Hash[maxn];
int bit[maxn];
void add(int pos, int val) {
    while(pos < n + 2) {
        bit[pos] = max(bit[pos], val);
        pos += lowbit(pos);
    }
}
int sum(int pos) {
    int res = 0;
    while(pos) {
        res = max(res, bit[pos]);
        pos -= lowbit(pos);
    }
    return res;
}
void init(int x) {
    for(int i = x; i < n + 2; i += lowbit(i))
        bit[i] = 0;
}
void solve(int l, int r) {

    int mid = (l + r) >> 1;
    sort(p + l, p + mid + 1, cmpy);
    sort(p + mid + 1, p + r + 1, cmpy);
    int j = l;
    for(int i = mid + 1; i <= r; i++) {
        while(j <= mid && p[j].y < p[i].y) {
            add(p[j].z, dp[p[j].x]);
            j++;
        }
        int tmp = sum(p[i].z - 1) + 1;
        dp[p[i].x] = max(dp[p[i].x], tmp);
    }
    for(int i = l; i <= mid; i++) init(p[i].z);
    sort(p + mid + 1, p + r + 1, cmpx);
}
void CDQ(int l, int r) {
    if(l == r) {
        return;
    }
    int mid = (l + r) >> 1;
    CDQ(l, mid);
    solve(l, r);
    CDQ(mid + 1, r);
}
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d%d", &p[i].y, &p[i].z);
        p[i].x = i;
        dp[i] = 1;
        Hash[i] = p[i].z;
    }
    
    sort(Hash + 1, Hash + n + 1);
    int cnt = unique(Hash + 1, Hash + n + 1) - Hash - 1;
    for(int i = 1; i <= n; i++) {
        p[i].z = lower_bound(Hash + 1, Hash + cnt + 1, p[i].z) - Hash;
    }
    // debug1(n);
    CDQ(1, n);
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        ans = max(ans, dp[i]);
    }
    printf("%d\n", ans);

    return 0;
}