bzoj 4553 && HEOI2016 day1t3 seq

Posted 2020-08-22 SD_le

      一个序列在所有变换中都单调不降的条件是i<j,a[i]<=min[j],mx[i]<=a[j],所以套CDQ就行了。

     

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<algorithm>
  4 #include<cstring>
  5 #define N 100005
  6 #define inf 0x3f3f3f3f
  7 using namespace std;
  8 int n,m;
  9 int a[N];
 10 int f[N];
 11 int mx[N],mn[N];
 12 int tmp[N];
 13 int v[N];
 14 int res;
 15 bool cmp(int x,int y)
 16 {
 17     if(x<=res&&y<=res)return mx[x]<mx[y];
 18     if(x<=res)return mx[x]<=a[y];
 19     if(y<=res)return a[x]<mx[y];
 20     return a[x]<a[y];
 21 }
 22 int dp[N];
 23 struct node
 24 {
 25     int lazy,zhi;
 26 }aa[N*8];
 27 void push_down(int x)
 28 {
 29     if(aa[x].lazy!=0)
 30     {
 31         aa[x*2].zhi=aa[x*2+1].zhi=-inf;
 32         aa[x*2].lazy=aa[x*2+1].lazy=1;
 33         aa[x].lazy=0;
 34     }
 35     return ;
 36 }
 37 int qur(int x,int l,int r,int ll,int rr)
 38 {
 39     if(l>=ll&&r<=rr)
 40     {
 41         return aa[x].zhi;
 42     }
 43     push_down(x);
 44     int mid=(l+r)>>1;
 45     if(rr<=mid)return qur(x*2,l,mid,ll,rr);
 46     if(ll>mid)return qur(x*2+1,mid+1,r,ll,rr);
 47     return max(qur(x*2,l,mid,ll,rr),qur(x*2+1,mid+1,r,ll,rr));
 48 }
 49 void gai(int x,int l,int r,int pos,int z)
 50 { 
 51      if(l==r)
 52      {
 53          aa[x].zhi=max(aa[x].zhi,z);return ;
 54      }
 55      push_down(x);
 56      int mid=(l+r)>>1;
 57      if(pos<=mid)gai(x*2,l,mid,pos,z);
 58      else gai(x*2+1,mid+1,r,pos,z);
 59      aa[x].zhi=max(aa[x*2].zhi,aa[x*2+1].zhi);
 60 } 
 61 void solve(int l,int r)
 62 { 
 63     if (l==r)
 64     { 
 65         f[l]=max(f[l],1);
 66         return ; 
 67     }
 68     int mid=(l+r)>>1;
 69     solve(l,mid);res=mid;
 70     int cnt=0;
 71     for(int i=l;i<=r;i++)tmp[++cnt]=i;
 72     sort(tmp+1,tmp+cnt+1,cmp);
 73     aa[1].lazy=1;aa[1].zhi=-inf;
 74     for(int i=1;i<=cnt;i++)
 75     {
 76         if(tmp[i]<=mid)gai(1,1,100000,a[tmp[i]],f[tmp[i]]);
 77         else f[tmp[i]]=max(f[tmp[i]],qur(1,1,100000,1,mn[tmp[i]])+1);
 78     }
 79     solve(mid+1,r);
 80 }
 81 int main()
 82 {
 83    scanf("%d%d",&n,&m);
 84    memset(mx,0xcf,sizeof(mx));
 85    memset(mn,0x3f,sizeof(mn));
 86    for(int i=1;i<=n;i++)
 87    {
 88           scanf("%d",&a[i]);
 89           mx[i]=mn[i]=a[i];
 90    }
 91    for(int i=1;i<=m;i++)
 92    {
 93            int t1,t2;
 94            scanf("%d%d",&t1,&t2);
 95            mx[t1]=max(mx[t1],t2);
 96            mn[t1]=min(mn[t1],t2);
 97    }
 98    solve(1,n);
 99    int ans=1;
100    for(int i=1;i<=n;i++)ans=max(ans,f[i]);
101    printf("%d\n",ans);
102    return 0;
103 }