BZOJ3744: Gty的妹子序列

By Autumn

题解：

出题人卡常数丧心病狂。。。

我原来思考过这个问题，还想什么时候出道题呢，结果已经被出出来了。。。

我们分块，然后预处理出i到j块的逆序对数f[i][j]，然后查询的时候不在块内的就暴力查询区间内<或>该数的个数，然后就可以n√n*logn了。。

然后出题人卡常数啊。。。

预处理的时候我偷懒用了主席树结果就T成翔啊。。。

无奈写了BIT。。。

代码：TLE

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 50000+5
14 #define maxm 1000000+5
15 #define maxk 500+5
16 #define eps 1e-10
17 #define ll long long
18 #define pa pair<int,int>
19 #define for0(i,n) for(int i=0;i<=(n);i++)
20 #define for1(i,n) for(int i=1;i<=(n);i++)
21 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
22 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
23 #define mod 1000000007
24 #define l(x) ((x-1)*block+1)
25 #define r(x) ((x)==b[n]?n:(x)*block)
26 using namespace std;
27 inline int read()
28 {
29     int x=0,f=1;char ch=getchar();
30     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
31     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
32     return x*f;
33 }
34 int n,m,tot,a[maxn],b[maxn],v[maxn],s[maxm],ls[maxm],rs[maxm],rt[maxn],f[maxk][maxk];
35 inline bool cmp1(int x,int y){return a[x]<a[y];}
36 inline void update(int l,int r,int x,int &y,int z)
37 {
38     y=++tot;
39     s[y]=s[x]+1;
40     if(l==r)return;
41     ls[y]=ls[x];rs[y]=rs[x];
42     int mid=(l+r)>>1;
43     if(z<=mid)update(l,mid,ls[x],ls[y],z);else update(mid+1,r,rs[x],rs[y],z);
44 }
45 inline int query(int x,int y,int z)
46 {
47     int l=1,r=m,mid,xx=rt[x-1],yy=rt[y],res=0;
48     while(l!=r)
49     {
50         mid=(l+r)>>1;
51         if(z<=mid){xx=ls[xx];yy=ls[yy];r=mid;}
52         else {res+=s[ls[yy]]-s[ls[xx]];xx=rs[xx];yy=rs[yy];l=mid+1;}
53     }
54     return res;
55 }
56 int main()
57 {
58     freopen("input.txt","r",stdin);
59     freopen("output.txt","w",stdout);
60     n=read();
61     for1(i,n)a[i]=read(),b[i]=i;
62     sort(b+1,b+n+1,cmp1);
63     for1(i,n)
64     {
65         if(i==1||(a[b[i]]!=a[b[i-1]]))m++;
66         v[b[i]]=m;
67     }
68     m++;
69     for1(i,n)update(1,m,rt[i-1],rt[i],v[i]);
70     int block=sqrt(n)*log(n)/log(2);
71     for1(i,n)b[i]=(i-1)/block+1;
72     for1(i,b[n])
73     {
74         int tmp=0;
75         for2(j,l(i),n)tmp+=(j-l(i)+1)-query(l(i),j,v[j]+1),f[i][b[j]]=tmp;
76     }
77     int cs=read(),ans=0;
78     while(cs--)
79     {
80         int x=read()^ans,y=read()^ans;if(x>y)swap(x,y);int bx=b[x],by=b[y];
81         if(by-bx<2)
82         {
83             ans=0;
84             for2(i,x,y-1)ans+=query(i,y,v[i]);
85         }
86         else
87         {
88         ans=f[bx][by];
89         for2(i,l(bx),x-1)ans-=query(i,r(by),v[i]);
90         for2(i,y+1,r(by))ans-=i-x+1-query(x,i,v[i]+1);
91         }
92         printf("%d\n",ans);
93     }
94     return 0;
95 }

View Code

17s

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cmath>
  4 #include<cstring>
  5 #include<algorithm>
  6 #include<iostream>
  7 #include<vector>
  8 #include<map>
  9 #include<set>
 10 #include<queue>
 11 #include<string>
 12 #define inf 1000000000
 13 #define maxn 50000+5
 14 #define maxm 3000000+5
 15 #define maxk 1000+5
 16 #define eps 1e-10
 17 #define ll long long
 18 #define pa pair<int,int>
 19 #define for0(i,n) for(int i=0;i<=(n);i++)
 20 #define for1(i,n) for(int i=1;i<=(n);i++)
 21 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 22 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 23 #define mod 1000000007
 24 #define l(x) ((x-1)*block+1)
 25 #define r(x) ((x)==b[n]?n:(x)*block)
 26 using namespace std;
 27 inline int read()
 28 {
 29     int x=0,f=1;char ch=getchar();
 30     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 31     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 32     return x*f;
 33 }
 34 int n,m,tot,a[maxn],b[maxn],v[maxn],s[maxm],ss[maxn],ls[maxm],rs[maxm],rt[maxn],f[maxk][maxk];
 35 inline bool cmp1(int x,int y){return a[x]<a[y];}
 36 inline void update(int l,int r,int x,int &y,int z)
 37 {
 38     y=++tot;
 39     s[y]=s[x]+1;
 40     if(l==r)return;
 41     ls[y]=ls[x];rs[y]=rs[x];
 42     int mid=(l+r)>>1;
 43     if(z<=mid)update(l,mid,ls[x],ls[y],z);else update(mid+1,r,rs[x],rs[y],z);
 44 }
 45 inline int query(int x,int y,int z)
 46 {
 47     int l=1,r=m,mid,xx=rt[x-1],yy=rt[y],res=0;
 48     while(l!=r)
 49     {
 50         mid=(l+r)>>1;
 51         if(z<=mid){xx=ls[xx];yy=ls[yy];r=mid;}
 52         else {res+=s[ls[yy]]-s[ls[xx]];xx=rs[xx];yy=rs[yy];l=mid+1;}
 53     }
 54     return res;
 55 }
 56 inline int sum(int x){int t=0;for(;x;x-=x&(-x))t+=ss[x];return t;}
 57 inline void add(int x,int y){for(;x<=m;x+=x&(-x))ss[x]+=y;} 
 58 int main()
 59 {
 60     freopen("input.txt","r",stdin);
 61     freopen("output.txt","w",stdout);
 62     n=read();
 63     for1(i,n)a[i]=read(),b[i]=i;
 64     sort(b+1,b+n+1,cmp1);
 65     for1(i,n)
 66     {
 67         if(i==1||(a[b[i]]!=a[b[i-1]]))m++;
 68         v[b[i]]=m;
 69     }
 70     m++;
 71     for1(i,n)update(1,m,rt[i-1],rt[i],v[i]);
 72     int block=sqrt(n);
 73     for1(i,n)b[i]=(i-1)/block+1;
 74     for1(i,b[n])
 75     {
 76         int tmp=0;
 77         for2(j,l(i),n)
 78         {
 79          add(v[j],1);
 80          tmp+=j-l(i)+1-sum(v[j]);
 81          f[i][b[j]]=tmp;
 82         }
 83         for2(j,l(i),n)add(v[j],-1);
 84     }
 85     //for1(i,b[n])for1(j,b[n])cout<<i<<‘ ‘<<j<<‘ ‘<<f[i][j]<<endl;
 86     int cs=read(),ans=0;
 87     while(cs--)
 88     {
 89         int x=read()^ans,y=read()^ans;if(x>y)swap(x,y);int bx=b[x],by=b[y];
 90         if(by-bx<2)
 91         {
 92             ans=0;
 93             for2(i,x,y-1)ans+=query(i,y,v[i]);
 94         }
 95         else
 96         {
 97         ans=f[bx][by];
 98         for2(i,l(bx),x-1)ans-=query(i,r(by),v[i]);
 99         for2(i,y+1,r(by))ans-=i-x+1-query(x,i,v[i]+1);
100         }
101         printf("%d\n",ans);
102     }
103     return 0;
104 }

View Code

20s

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cmath>
  4 #include<cstring>
  5 #include<algorithm>
  6 #include<iostream>
  7 #include<vector>
  8 #include<map>
  9 #include<set>
 10 #include<queue>
 11 #include<string>
 12 #define inf 1000000000
 13 #define maxn 50000+5
 14 #define maxm 3000000+5
 15 #define maxk 1000+5
 16 #define eps 1e-10
 17 #define ll long long
 18 #define pa pair<int,int>
 19 #define for0(i,n) for(int i=0;i<=(n);i++)
 20 #define for1(i,n) for(int i=1;i<=(n);i++)
 21 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 22 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 23 #define mod 1000000007
 24 #define l(x) ((x-1)*block+1)
 25 #define r(x) ((x)==b[n]?n:(x)*block)
 26 using namespace std;
 27 inline int read()
 28 {
 29     int x=0,f=1;char ch=getchar();
 30     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 31     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 32     return x*f;
 33 }
 34 int n,m,tot,a[maxn],b[maxn],v[maxn],s[maxm],ss[maxn],ls[maxm],rs[maxm],rt[maxn],f[maxk][maxk];
 35 inline bool cmp1(int x,int y){return a[x]<a[y];}
 36 inline void update(int l,int r,int x,int &y,int z)
 37 {
 38     y=++tot;
 39     s[y]=s[x]+1;
 40     if(l==r)return;
 41     ls[y]=ls[x];rs[y]=rs[x];
 42     int mid=(l+r)>>1;
 43     if(z<=mid)update(l,mid,ls[x],ls[y],z);else update(mid+1,r,rs[x],rs[y],z);
 44 }
 45 inline int query(int x,int y,int z)
 46 {
 47     int l=1,r=m,mid,xx=rt[x-1],yy=rt[y],res=0;
 48     while(l!=r)
 49     {
 50         mid=(l+r)>>1;
 51         if(z<=mid){xx=ls[xx];yy=ls[yy];r=mid;}
 52         else {res+=s[ls[yy]]-s[ls[xx]];xx=rs[xx];yy=rs[yy];l=mid+1;}
 53     }
 54     return res;
 55 }
 56 inline int sum(int x){int t=0;for(;x;x-=x&(-x))t+=ss[x];return t;}
 57 inline void add(int x,int y){for(;x<=m;x+=x&(-x))ss[x]+=y;} 
 58 int main()
 59 {
 60     freopen("input.txt","r",stdin);
 61     freopen("output.txt","w",stdout);
 62     n=read();
 63     for1(i,n)a[i]=read(),b[i]=i;
 64     sort(b+1,b+n+1,cmp1);
 65     for1(i,n)
 66     {
 67         if(i==1||(a[b[i]]!=a[b[i-1]]))m++;
 68         v[b[i]]=m;
 69     }
 70     m++;
 71     for1(i,n)update(1,m,rt[i-1],rt[i],v[i]);
 72     int block=sqrt(n);
 73     for1(i,n)b[i]=(i-1)/block+1;
 74     for1(i,b[n])
 75     {
 76         int tmp=0;
 77         for2(j,l(i),n)
 78         {
 79          add(v[j],1);
 80          tmp+=j-l(i)+1-sum(v[j]);
 81          f[i][b[j]]=tmp;
 82         }
 83         for2(j,l(i),n)add(v[j],-1);
 84     }
 85     //for1(i,b[n])for1(j,b[n])cout<<i<<‘ ‘<<j<<‘ ‘<<f[i][j]<<endl;
 86     int cs=read(),ans=0;
 87     while(cs--)
 88     {
 89         int x=read()^ans,y=read()^ans;if(x>y)swap(x,y);int bx=b[x],by=b[y];
 90         if(by-bx<2)
 91         {
 92             ans=0;
 93             for2(i,x,y-1)ans+=query(i,y,v[i]);
 94         }
 95         else
 96         {
 97         ans=f[bx+1][by-1];
 98         for2(i,x,r(bx))ans+=query(i,r(by-1),v[i]);
 99         for2(i,l(by),y)ans+=i-x+1-query(x,i,v[i]+1);
100         }
101         printf("%d\n",ans);
102     }
103     return 0;
104 }

View Code

BZOJ3744: Gty的妹子序列

3744: Gty的妹子序列

Description

Input

Output

Sample Input

Sample Output

HINT

Source