bzoj2244[SDOI2011]拦截导弹

http://www.lydsy.com/JudgeOnline/problem.php?id=2244

第$i$个导弹看成一个三元组$(i,h_i,v_i)$

其实就是最长上升子序列的问题。

我们分别求以第$i$个导弹为结尾的最长上升子序列的长度和个数，以及以第$i$个导弹为开头的最长上升子序列的长度和个数。

下面以求以第$i$个导弹为结尾的最长上升子序列的长度和个数为例。

记以第$i$个导弹结尾的最长上升子序列长度为$f[i]$，则：

$$f[i]=Max\{f[j]|j<i,h[j]\geq h[i],v[j]\geq v[i]\}+1$$

所以第1问就是三维偏序，用CDQ分治解决。

第一维由分治的时候左边小于右边保证；

第二维由排序保证；

第三维由数据结构保证；

看程序应该会比较好理解。

记以第$i$个导弹结尾的最长上升子序列长度为$g[i]$，则：

$$g[i]=\sum g[j](f[j]+1=f[i],j<i,h[j]\geq h[i],v[j]\geq v[i])$$

我们可以先按f排序，然后分层DP，还是用CDQ分治。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj

using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define fill(a,l,r,v) fill(a+l,a+r+1,v)
#define re(i,a,b)  for(i=(a);i<=(b);i++)
#define red(i,a,b) for(i=(a);i>=(b);i--)
#define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define p_b(a) push_back(a)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z==‘-‘){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar());
        return (neg)?-res:res;
    }
LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z==‘-‘){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar());
        return (neg)?-res:res;
    }

const int maxn=50000;

int n,ny,nz;
struct Ta{int x,y,z,f,flag;DB g;}a[maxn+100];
int bak[maxn+100];

int f1[maxn+100],f2[maxn+100];
DB g1[maxn+100],g2[maxn+100];
int ans;

bool cmpx(Ta a,Ta b){return a.x<b.x;}
bool cmpyx(Ta a,Ta b){return a.y!=b.y?a.y<b.y:a.x<b.x;}//注意y相同时，x小的在前面！！！
bool cmpf(Ta a,Ta b){return a.f<b.f;}

#define lowbit(a) ((a)&(-a))
namespace TREE1
  {
        int tree[maxn+100];
        void update(int a,int v){for(;a<=n;a+=lowbit(a))upmax(tree[a],v);}
        int ask(int a){int res=0;for(;a>=1;a-=lowbit(a))upmax(res,tree[a]);return res;}
        void clear(int a){for(;a<=n;a+=lowbit(a))tree[a]=0;}
  }
void CDQf(int l,int r)
  {
      if(l==r){upmax(a[l].f,1);return;}
      int i,mid=(l+r)/2;
      CDQf(l,mid);
      sort(a+l,a+r+1,cmpyx);
      re(i,l,r)if(a[i].x<=mid)TREE1::update(a[i].z,a[i].f);else upmax(a[i].f,TREE1::ask(a[i].z)+1);
      re(i,l,r)if(a[i].x<=mid)TREE1::clear(a[i].z);
      sort(a+l,a+r+1,cmpx);
      CDQf(mid+1,r);
  }

namespace TREE2
  {
      DB tree[maxn+100];
      void update(int a,DB v){for(;a<=n;a+=lowbit(a))tree[a]+=v;}
      DB ask(int a){DB res=0;for(;a>=1;a-=lowbit(a))res+=tree[a];return res;}
      void clear(int a){for(;a<=n;a+=lowbit(a))tree[a]=0.0;}
  }
void CDQg(int l,int r)
  {
      if(l==r)return;
      int i,mid=(l+r)/2;
      CDQg(l,mid);
      sort(a+l,a+r+1,cmpyx);
      re(i,l,r)
        {
            if(a[i].x<=mid && a[i].flag) TREE2::update(a[i].z,a[i].g);
            if(a[i].x>mid && !a[i].flag) a[i].g+=TREE2::ask(a[i].z);
        }
      re(i,l,r)if(a[i].x<=mid && a[i].flag) TREE2::clear(a[i].z);
      sort(a+l,a+r+1,cmpx);
      CDQg(mid+1,r);
  }

int tmpx[maxn+100];
void solve()
  {
      int i,j,l1,r1,l2,r2;
      re(i,1,n)a[i].f=0,a[i].g=0.0;
      CDQf(1,n);
      sort(a+1,a+n+1,cmpf);
      for(l1=r1=1;r1<n && a[r1+1].f==a[l1].f;r1++);
      re(i,l1,r1)a[i].g=1;
      for(;r1<n;l1=l2,r1=r2)
        {
            for(l2=r2=r1+1;r2<n && a[r2+1].f==a[l2].f;r2++);
            re(i,l1,r1)a[i].flag=1;
            sort(a+l1,a+r2+1,cmpx);
            re(i,l1,r2)tmpx[i]=a[i].x,a[i].x=i;
            CDQg(l1,r2);
            re(i,l1,r2)a[i].x=tmpx[i];
            sort(a+l1,a+r2+1,cmpf);
            re(i,l1,r1)a[i].flag=0;
        }
      sort(a+1,a+n+1,cmpx);
  }

int ans1;
DB ans2;

int main()
  {
      freopen("bzoj2244.in","r",stdin);
      freopen("bzoj2244.out","w",stdout);
        int i;
        n=gint();
        re(i,1,n)a[i].x=i,a[i].y=gint(),a[i].z=gint();

        ny=0;
        re(i,1,n)bak[++ny]=a[i].y;
        sort(bak+1,bak+ny+1);
        ny=unique(bak+1,bak+ny+1)-bak-1;
        re(i,1,n)a[i].y=ny-(lower_bound(bak+1,bak+ny+1,a[i].y)-bak)+1;
        nz=0;
        re(i,1,n)bak[++nz]=a[i].z;
        sort(bak+1,bak+nz+1);
        nz=unique(bak+1,bak+nz+1)-bak-1;
        re(i,1,n)a[i].z=nz-(lower_bound(bak+1,bak+nz+1,a[i].z)-bak)+1;

        solve();
        re(i,1,n)f1[i]=a[i].f,g1[i]=a[i].g;

        re(i,1,n)a[i].x=n-a[i].x+1,a[i].y=ny-a[i].y+1,a[i].z=nz-a[i].z+1;
        re(i,1,n/2)swap(a[i],a[n-i+1]);

        solve();
        re(i,1,n)a[i].x=n-a[i].x+1,a[i].y=ny-a[i].y+1,a[i].z=nz-a[i].z+1;
        re(i,1,n/2)swap(a[i],a[n-i+1]);
        re(i,1,n)f2[i]=a[i].f,g2[i]=a[i].g;

        re(i,1,n)upmax(ans1,f1[i]);
        re(i,1,n)if(f1[i]==ans1)ans2+=g1[i];
        PF("%d\n",ans1);
        re(i,1,n)if(f1[i]+f2[i]-1!=ans1 || sgn(ans2)==0)PF("0.00000 ");else PF("%0.5lf ",g1[i]*g2[i]/ans2);
        return 0;
    }