BZOJ - 2244 拦截导弹 (dp,CDQ分治+树状数组优化)

BZOJ - 2244 拦截导弹 (dp,CDQ分治+树状数组优化)

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #define MAXN 50010
 6 #define LL long long
 7 using namespace std;
 8 struct ddd
 9 {    
10     int t,h,v,ans;    
11     #define t(x) a[x].t
12     #define h(x) a[x].h    
13     #define v(x) a[x].v
14     #define ans(x) a[x].ans
15 }a[MAXN];
16 int n,th[MAXN],tv[MAXN];
17 bool cmp1(ddd a,ddd b){return a.t==b.t?(a.h==b.h?(a.v<b.v):a.h<b.h):a.t<b.t;}
18 bool cmp2(ddd a,ddd b){return a.h==b.h?(a.t==b.t?(a.v>b.v):a.t<b.t):a.h>b.h;}
19 int ans=1;
20 int C[MAXN];
21 #define lowbit(x) ((x)&(-(x)))
22 //void add(int x,int y){while(x<=n){C[x]+=y;x+=lowbit(x);}}
23 //int ask(int x){int ans=0;while(x){ans+=C[x];x-=lowbit(x);}return ans;}
24 void add(int x,int y)
25 {//cout<<"add "<<x<<" "<<y<<endl;
26 while(x){C[x]=max(C[x],y);x-=lowbit(x);}}
27 void admin(int x,int y)
28 {while(x){C[x]=min(C[x],y);x-=lowbit(x);}}
29 int ask(int x){int ans=0;while(x<=n){ans=max(ans,C[x]);x+=lowbit(x);}return ans;}
30 void CDQ(int l,int r)
31 {    
32     if(l==r){return;}
33     int mid=(l+r)>>1;
34 //    sort(a+l,a+mid+1,cmp1);
35     CDQ(l,mid);
36     sort(a+l,a+mid+1,cmp2);
37     sort(a+mid+1,a+r+1,cmp2);
38     int j=l;
39 //    cout<<"# "<<l<<" "<<r<<endl;
40 //    for(int i=l;i<=r;i++)cout<<i<<": "<<t(i)<<" "<<h(i)<<" "<<v(i)<<endl;
41     for(int i=mid+1;i<=r;i++)
42     {
43         while(j<=mid&&h(j)>=h(i)){//cout<<"!!!!!!!!j "<<j<<" "<<ans(j)<<endl;
44         add(v(j),ans(j)),j++;}
45         ans(i)=max(ans(i),ask(v(i))+1);//cout<<"!!!!!!!!!!ans "<<i<<" "<<ans(i)<<endl;
46     }    
47     for(int i=l;i<j;i++)admin(v(i),0);
48     sort(a+mid+1,a+r+1,cmp1);
49     CDQ(mid+1,r);    
50 }
51 signed main()
52 {
53 //    freopen("in.txt","r",stdin);
54 
55     cin>>n;
56     for(int i=1;i<=n;i++)    
57         cin>>h(i)>>v(i),t(i)=i,th[i]=h(i),tv[i]=v(i);
58     sort(th+1,th+n+1);
59     sort(tv+1,tv+n+1);
60     int len1=unique(th+1,th+n+1)-th-1;
61     int len2=unique(tv+1,tv+n+1)-tv-1;
62     for(int i=1;i<=n;i++)
63         h(i)=lower_bound(th+1,th+len1+1,h(i))-th,
64         v(i)=lower_bound(tv+1,tv+len2+1,v(i))-tv;
65     sort(a+1,a+n+1,cmp1);
66     for(int i=1;i<=n;i++)ans(i)=1;
67     CDQ(1,n);
68     sort(a+1,a+n+1,cmp1);
69 //    for(int i=1;i<=n;i++)cout<<"ans "<<i<<" "<<ans(i)<<endl;
70     for(int i=1;i<=n;i++)ans=max(ans,ans(i));
71     printf("%d\n",ans);
72     for(int i=1;i<=n;i++) cout<<(double)1/3<<" ";puts("");
73 }
第一问

先说第一问吧，本来觉得挺简单的，做起来才发现一直调不出来……第一问其实是一个比较容易看出来的三位偏序问题，只不过并不是简单的求和，而是用一个点前面h和v都大于他的点答案的最大值+1去更新这个点，所以树状数组要维护后缀，记得要把每个点的初始答案附成1。

第二问就比较麻烦了，一个点出现的概率等于$\frac{包含这个点的方案数}{总方案数}$，维护4个数组，f[i]表示以i为结尾的最大长度，g[i]表示以i为结尾最大长度的方案数，h[i]表示以i开头的最大长度，s[i]表示以i开头最大长度的方案数。两个要跑两边CDQ，这里之说第一遍，用树状数组维护后缀最大值及其方案数（用结构体或pair实现），CDQ时先递归处理左区间，然后考虑当前区间，最后递归处理右区间。至于答案的更新：

		if(tem.maxn==f(i))g(i)+=tem.num;
		else if(tem.maxn>f(i))f(i)=tem.maxn,g(i)=tem.num;

 还有几点要注意：如果一个点不能包含在最长子序列中，那么选中他的概率为0。记得开double，这道题会爆longlong。

  1 #include<algorithm>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<cmath>
  6 #define MAXN 50010
  7 #define LL long long
  8 #define int LL
  9 using namespace std;
 10 struct ddd
 11 {    
 12     int t,h,v,ans;
 13     double f,g,he,s;    
 14     #define t(x) a[x].t
 15     #define h(x) a[x].h    
 16     #define v(x) a[x].v
 17     #define ans(x) a[x].ans
 18     #define f(x) a[x].f
 19     #define g(x) a[x].g
 20     #define he(x) a[x].he
 21     #define s(x) a[x].s
 22 }a[MAXN];
 23 int n,th[MAXN],tv[MAXN];
 24 bool cmp1(ddd a,ddd b){return a.t==b.t?(a.h==b.h?(a.v<b.v):a.h<b.h):a.t<b.t;}
 25 bool cmp2(ddd a,ddd b){return a.h==b.h?(a.t==b.t?(a.v>b.v):a.t<b.t):a.h>b.h;}
 26 bool cmp3(ddd a,ddd b){return a.t==b.t?(a.h==b.h?(a.v<b.v):a.h<b.h):a.t>b.t;}
 27 bool cmp4(ddd a,ddd b){return a.h==b.h?(a.t==b.t?(a.v<b.v):a.t<b.t):a.h<b.h;}
 28 double ans=1;
 29 #define lowbit(x) ((x)&(-(x)))
 30 struct cc{double maxn;double num;}C[MAXN];
 31 void add(int x,double y,double nu)
 32 {
 33     while(x)
 34     {
 35         if(C[x].maxn==y)C[x].num+=nu;
 36         else if(C[x].maxn<y)C[x].maxn=y,C[x].num=nu;
 37         x-=lowbit(x);
 38     }
 39 }
 40 void admin(int x)
 41 {while(x){C[x].maxn=C[x].num=0;x-=lowbit(x);}}
 42 cc ask(int x)
 43 {
 44     cc ans={0,0};
 45     while(x<=n)
 46     {
 47         if(C[x].maxn==ans.maxn)ans.num+=C[x].num;
 48         else if(C[x].maxn>ans.maxn)ans.maxn=C[x].maxn,ans.num=C[x].num;
 49         x+=lowbit(x);
 50     }
 51     return ans;
 52 }
 53 cc C2[MAXN];
 54 void add2(int x,double y,double nu)
 55 {
 56     while(x<=n)
 57     {
 58         if(C2[x].maxn==y)C2[x].num+=nu;
 59         else if(C2[x].maxn<y)C2[x].maxn=y,C2[x].num=nu;
 60         x+=lowbit(x);
 61     }
 62 }
 63 void admin2(int x)
 64 {while(x<=n){C2[x].maxn=C2[x].num=0;x+=lowbit(x);}}
 65 cc ask2(int x)
 66 {
 67     cc ans={0,0};
 68     while(x)
 69     {
 70         if(C2[x].maxn==ans.maxn)ans.num+=C2[x].num;
 71         else if(C2[x].maxn>ans.maxn)ans.maxn=C2[x].maxn,ans.num=C2[x].num;
 72         x-=lowbit(x);
 73     }
 74     return ans;
 75 }
 76 void CDQ(int l,int r)
 77 {    
 78     if(l==r){return;}
 79     int mid=(l+r)>>1;
 80     CDQ(l,mid);
 81     sort(a+l,a+mid+1,cmp2);
 82     sort(a+mid+1,a+r+1,cmp2);
 83     int j=l;
 84     for(int i=mid+1;i<=r;i++)
 85     {
 86         while(j<=mid&&h(j)>=h(i))
 87         {
 88             add(v(j),f(j),g(j));j++;
 89         }
 90         cc tem=ask(v(i));tem.maxn++;
 91         if(tem.maxn==f(i))g(i)+=tem.num;
 92         else if(tem.maxn>f(i))f(i)=tem.maxn,g(i)=tem.num;
 93     }    
 94     for(int i=l;i<j;i++)admin(v(i));
 95     sort(a+mid+1,a+r+1,cmp1);
 96     CDQ(mid+1,r);    
 97 }
 98 void CDQ2(int l,int r)
 99 {    
100     if(l==r){return;}
101     int mid=(l+r)>>1;
102     CDQ2(l,mid);
103     sort(a+l,a+mid+1,cmp4);
104     sort(a+mid+1,a+r+1,cmp4);
105     int j=l;
106     for(int i=mid+1;i<=r;i++)
107     {
108         while(j<=mid&&h(j)<=h(i))
109         {
110             add2(v(j),he(j),s(j));j++;
111         }
112         cc tem=ask2(v(i));tem.maxn++;
113         if(tem.maxn==he(i))s(i)+=tem.num;
114         else if(tem.maxn>he(i))he(i)=tem.maxn,s(i)=tem.num;
115     }    
116     for(int i=l;i<j;i++)admin2(v(i));
117     sort(a+mid+1,a+r+1,cmp3);
118     CDQ2(mid+1,r);    
119 }
120 signed main()
121 {
122 //    freopen("1.in","r",stdin);
123 //    freopen("out.out","w",stdout);
124 
125     cin>>n;
126     for(int i=1;i<=n;i++)    
127         cin>>h(i)>>v(i),t(i)=i,th[i]=h(i),tv[i]=v(i);
128     sort(th+1,th+n+1);
129     sort(tv+1,tv+n+1);
130     int len1=unique(th+1,th+n+1)-th-1;
131     int len2=unique(tv+1,tv+n+1)-tv-1;
132     for(int i=1;i<=n;i++)
133         h(i)=lower_bound(th+1,th+len1+1,h(i))-th,
134         v(i)=lower_bound(tv+1,tv+len2+1,v(i))-tv;
135     for(int i=1;i<=n;i++)f(i)=g(i)=he(i)=s(i)=1;
136     sort(a+1,a+n+1,cmp1);
137     CDQ(1,n);
138      for(int i=1;i<=n;i++)ans=max(ans,f(i));
139     printf("%0.0lf\n",ans);
140     sort(a+1,a+n+1,cmp3);
141     CDQ2(1,n);
142     double summ=0;
143     sort(a+1,a+n+1,cmp1);
144     for(int i=1;i<=n;i++)
145         if(fabs(f(i)-ans)<=1e-8) summ+=g(i);
146     for(int i=1;i<=n;i++)
147     {
148         if(f(i)+he(i)-1==ans){double ttt=(double)(1.0*g(i)*s(i))/summ;printf("%0.5lf ",ttt);}
149         else printf("0.00000 ");
150     }puts("");
151 }
View Code

 

 

 

posted @ 2019-08-03 06:11  Al_Ca  阅读(173)  评论(0)  编辑  收藏  举报
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