4990: [Usaco2017 Feb]Why Did the Cow Cross the Road II 线段树维护dp

题目 4990: [Usaco2017 Feb]Why Did the Cow Cross the Road II

连接

http://www.lydsy.com/JudgeOnline/problem.php?id=4990

题面

上下有两个长度为n、位置对应的序列A、B，
其中数的范围均为1~n。若abs(A[i]-B[j]) <= 4，
则A[i]与B[j]间能够连一条边。现要求在边与边不相交的状况下的最大的连边数量。
n <= 10^5。

输入

The first line of input contains N (1≤N≤100,0000).
The next N lines describe the order, by breed ID, of fields on one side of the road;
each breed ID is an integer in the range 1…N
The last N lines describe the order, by breed ID, of the fields on the other side of the road.
Each breed ID appears exactly once in each ordering.

输出

Please output the maximum number of disjoint "friendly crosswalks" Farmer John can draw across the road.

样例输入

6

1

2

3

4

5

6

6

5

4

3

2

1

样例输出

5

题解

考虑二维dp[i][j]表示考虑到<a[i]，b[j]>的最大数目
dp[i][j]=max(dp[i-1][j],dp[i][j-1])
若是ai和bj可以链接,dp[i][j]=max(dp[i-1][j-1]+1)

显然第一维能够优化，而后用个线段树求前缀最大值便可。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
int a[maxn],b[maxn],c[maxn],n,dp[maxn];
struct node{int l,r,x;}t[maxn*4];
void build(int x,int l,int r)
{
    t[x].l=l,t[x].r=r;
    if(l==r){t[x].x=0;return;}
    int mid=(l+r)/2;
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
    t[x].x=max(t[x<<1].x,t[x<<1|1].x);
}
void update(int x,int l,int r,int v){
    int L=t[x].l,R=t[x].r;
    if(l==L&&R==r){
        t[x].x=v;
        return;
    }
    int mid=(L+R)/2;
    if(mid>=l)update(x<<1,l,r,v);
    if(mid<r)update(x<<1|1,l,r,v);
    t[x].x=max(t[x<<1].x,t[x<<1|1].x);
}
int query(int x,int l,int r)
{
    int L=t[x].l,R=t[x].r;
    if(l<=L&&R<=r)return t[x].x;
    int ans=0,mid=(L+R)/2;
    if(mid>=l)ans=max(ans,query(x<<1,l,r));
    if(mid<r)ans=max(ans,query(x<<1|1,l,r));
    return ans;
}
int main(){
    scanf("%d",&n);
    build(1,1,n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&b[i]);
    }
    for(int i=1;i<=n;i++){
        c[b[i]]=i;
    }
    for(int i=1;i<=n;i++){
        for(int j=-4;j<=4;j++){
            if(a[i]+j>0&&a[i]+j<=n){
                dp[c[a[i]+j]]=max(dp[c[a[i]+j]],query(1,1,c[a[i]+j]-1)+1);
            }
        }
        for(int j=-4;j<=4;j++){
            if(a[i]+j>0&&a[i]+j<=n){
                update(1,c[a[i]+j],c[a[i]+j],dp[c[a[i]+j]]);
            }           
        }
    }
    cout<<query(1,1,n)<<endl;
}