【牛客网】Longest Common Subsequence

【牛客网】Longest Common Subsequence

发现只有d数组最格路

于是我们把前三个数组中相同的数记成一个三维坐标，同一个数坐标不会超过8个

从前往后枚举d，每次最多只会更新不超过8个点

而每个点更新就是找这个点三维偏序都小于它的最大的一个值+1来更新它

用KD树来维护，这个点与树中节点三维的某一维不相交就退出

可以加的剪枝是如果树中最大值+1小于当前搜到的答案就退出

然后把新的值在树中进行更新

跑的还是挺快的，0.5s不到

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 10005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,dimension;
int a[MAXN],b[MAXN],c[MAXN],d[MAXN],rt,Ncnt;
vector<int> v[3][MAXN];
struct node {
    int x,y,z;
    friend bool operator < (const node &a,const node &b) {
	if(dimension == 0) return a.x < b.x;
	else if(dimension == 1) return a.y < b.y;
	else return a.z < b.z;
    }
    friend bool operator == (const node &a,const node &b) {
	return a.x == b.x && a.y == b.y && a.z == b.z;
    }
    friend void upmin(node &a,node b) {
	a.x = min(a.x,b.x);a.y = min(a.y,b.y);a.z = min(a.z,b.z);
    }
    friend void upmax(node &a,node b) {
	a.x = max(a.x,b.x);a.y = max(a.y,b.y);a.z = max(a.z,b.z);
    }
};
struct KD_node {
    node p;int v,maxv;
    node av,bv;
    int lc,rc;
}tr[MAXN * 8];
vector<node> pos[MAXN],line;
vector<int> rec[MAXN];
bool vis[15];
#define lc(u) tr[u].lc
#define rc(u) tr[u].rc
void build(int &u,int l,int r,int d) {
    if(l > r) return;
    u = ++Ncnt;
    dimension = d;
    int mid = (l + r) >> 1;
    nth_element(line.begin() + l,line.begin() + mid,line.begin() + r + 1);
    tr[u].p = line[mid];tr[u].v = 0;tr[u].maxv = 0;
    tr[u].av = line[mid];tr[u].bv = line[mid];
    build(lc(u),l,mid - 1,(d + 1) % 3);
    build(rc(u),mid + 1,r,(d + 1) % 3);
    upmin(tr[u].av,tr[lc(u)].av);upmin(tr[u].av,tr[rc(u)].av);
    upmax(tr[u].bv,tr[lc(u)].bv);upmax(tr[u].bv,tr[rc(u)].bv);
}
bool sless(node a,node b) {
    return a.x < b.x && a.y < b.y && a.z < b.z;
}
bool uless(node a,node b) {
    return a.x <= b.x && a.y <= b.y && a.z <= b.z;
}
bool checkin(node x,int u) {
    return uless(tr[u].av,x) && uless(x,tr[u].bv);
}
void update(int u) {
    tr[u].maxv = max(tr[u].v,max(tr[lc(u)].maxv,tr[rc(u)].maxv));
}
void Change(int u,node p,int v) {
    if(!u) return;
    if(tr[u].p == p) {tr[u].v = v;update(u);return;}
    if(!checkin(p,u)) return;
    Change(lc(u),p,v);
    Change(rc(u),p,v);
    update(u);
}
int ta;
void Query(int u,node p) {
    if(tr[u].maxv + 1 <= ta) return;
    if(!sless(tr[u].av,p)) return;
    if(sless(tr[u].p,p)) ta = max(ta,tr[u].v + 1);
    Query(lc(u),p);Query(rc(u),p);
}
void Update_array(node p,int v) {
    ta = max(v,1);
    Query(1,p);
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);v[0][a[i]].pb(i);}
    for(int i = 1 ; i <= N ; ++i) {read(b[i]);v[1][b[i]].pb(i);}
    for(int i = 1 ; i <= N ; ++i) {read(c[i]);v[2][c[i]].pb(i);}
    for(int i = 1 ; i <= N ; ++i) read(d[i]);
    for(int i = 1 ; i <= N ; ++i) {
	for(auto t0 : v[0][i]) {
	    for(auto t1 : v[1][i]) {
		for(auto t2 : v[2][i]) {
		    pos[i].pb((node){t0,t1,t2});rec[i].pb(0);
		    line.pb((node){t0,t1,t2});
		}
	    }
	}
    }
    tr[0].av = (node){0x7fffffff,0x7fffffff,0x7fffffff};tr[0].bv = (node){-1,-1,-1};
    build(rt,0,line.size() - 1,0);
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
	memset(vis,0,sizeof(vis));
	for(int j = 0 ; j < pos[d[i]].size() ; ++j) {
	    Update_array(pos[d[i]][j],rec[d[i]][j]);
	    if(ta > rec[d[i]][j]) vis[j] = 1;
	    rec[d[i]][j] = ta;
	    ans = max(ans,ta);
	}
	for(int j = 0 ; j < pos[d[i]].size() ; ++j) {
	    if(vis[j]) Change(1,pos[d[i]][j],rec[d[i]][j]);
	}
    }
    out(ans);enter;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}