pair-pair

输入文件：pair-pair.in   输出文件：pair-pair.out   简单对比
时间限制：7 s  
内存限制：64 MB

Time Limit : 7000 MS

Memory Limit : 65536 KB

Pair-Pair

Bobo is tired of all kinds of hard LIS (Longest Increasing Subsequence) problems, so he decides to make himself some easier one.

Bobo has n pairs (a1,b1),(a2,b2),…,(an,bn) where 1≤ai,bi≤m holds for all
i. He defines f(i,j) be the length of longest increasing subsequence of
sequence {ai,bi,aj,bj}.

It‘s clear that 1≤f(i,j)≤4. Bobo would like to know g(k) which is the number of pairs (i,j) where f(i,j)=k.

Note that a sequence labeled with {i1,i2,…,ik} is an increasing subsequence of {a1,a2,…,an} only if:

1≤i1<i2<?<ik≤nai1<ai2<?<aik

Input

The first line contains 2 integers n,m (1≤n≤105,1≤m≤103).

The i-th of the following n lines contains 2 integers ai,bi (1≤ai,bi≤m).

Output

For each set, 4 integers g(1),g(2),g(3),g(4).

Sample Input

2 4

1 2

3 4

2 1

1 1

1 1

Sample Output

0 3 0 1

4 0 0 0

　　注意各种特判就好了。

　　有时间再更新题解吧……

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstring>
  4 #include <cstdio>
  5 using namespace std;
  6 const int maxn=100010;
  7 const int maxm=1010;
  8 int n,m;
  9 long long a[10];
 10 long long b1[maxm],b2[maxm];
 11 long long b3[maxm],b4[maxm];
 12
 13 struct Node{
 14     int a,b;
 15     Node(int a_=0,int b_=0){
 16         a=a_;b=b_;
 17     }
 18 }p[maxn];
 19
 20 bool cmp(Node x,Node y){
 21     if(x.a!=y.a)
 22     return x.a<y.a;
 23     return x.b<y.b;
 24 }
 25
 26 void Bit_Add(long long *b,int x,int d){
 27     while(x<=m){
 28         b[x]+=d;
 29         x+=x&(-x);
 30     }
 31 }
 32
 33 int Bit_Query(long long *b,int x){
 34     int ret=0;
 35     while(x){
 36         ret+=b[x];
 37         x-=x&(-x);
 38     }
 39     return ret;
 40 }
 41
 42 int rt[maxm],sum[maxn*20],ch[maxn*20][2],cnt;
 43 void Insert(int pre,int &rt,int l,int r,int g,int d){
 44     rt=++cnt;
 45     ch[rt][0]=ch[pre][0];
 46     ch[rt][1]=ch[pre][1];
 47     sum[rt]=sum[pre]+d;
 48     if(l==r)return;
 49     int mid=(l+r)>>1;
 50     if(mid>=g)Insert(ch[pre][0],ch[rt][0],l,mid,g,d);
 51     else Insert(ch[pre][1],ch[rt][1],mid+1,r,g,d);
 52 }
 53
 54 int Query(int pre,int rt,int l,int r,int a,int b){
 55     if(a>b)return 0;
 56     if(l>=a&&r<=b)return sum[rt]-sum[pre];
 57     int mid=(l+r)>>1,ret=0;
 58     if(mid>=a)ret=Query(ch[pre][0],ch[rt][0],l,mid,a,b);
 59     if(mid<b)ret+=Query(ch[pre][1],ch[rt][1],mid+1,r,a,b);
 60     return ret;
 61 }
 62
 63 void Init(){
 64     memset(a,0,sizeof(a));cnt=0;
 65     memset(b1,0,sizeof(b1));
 66     memset(b2,0,sizeof(b2));
 67     memset(b3,0,sizeof(b3));
 68     memset(b4,0,sizeof(b4));
 69 }
 70
 71 int main(){
 72 #ifndef ONLINE_JUDGE
 73     freopen("pair-pair.in","r",stdin);
 74     freopen("pair-pair.out","w",stdout);
 75 #endif
 76     while(scanf("%d%d",&n,&m)!=EOF){
 77         Init();
 78         for(int i=1;i<=n;i++)
 79             scanf("%d%d",&p[i].a,&p[i].b);
 80         sort(p+1,p+n+1,cmp);
 81         for(int i=1,last=0;i<=n;i++){
 82             long long tot=2*(i-1),tmp;
 83
 84             if(p[i].a<p[i].b){
 85                 tmp=Bit_Query(b1,m)-Bit_Query(b1,p[i].b)+Bit_Query(b2,p[i].a-1);
 86                 a[4]+=tmp;tot-=tmp;
 87
 88                 tmp=Bit_Query(b1,p[i].b)-Bit_Query(b1,p[i].a);
 89                 tmp+=Bit_Query(b2,p[i].b-1)-Bit_Query(b2,p[i].a-1);
 90
 91                 tmp+=Bit_Query(b3,m)-Bit_Query(b3,p[i].b)+Bit_Query(b4,p[i].a-1);
 92                 tmp+=Bit_Query(b2,m)-Bit_Query(b2,p[i].b);
 93                 for(int j=last+1;j<p[i].a;j++)rt[j]=rt[last];
 94
 95                 tmp+=Query(rt[0],rt[p[i].a-1],1,m,p[i].b,m);
 96
 97                 a[3]+=tmp;tot-=tmp;
 98
 99                 Insert(rt[last],rt[p[i].a],1,m,p[i].b,1);
100
101                 last=p[i].a;a[2]+=tot;
102
103                 Bit_Add(b1,p[i].a,1);Bit_Add(b2,p[i].b,1);
104             }
105             else{
106                 tmp=Bit_Query(b3,p[i].b)+Bit_Query(b4,m)-Bit_Query(b4,p[i].a-1);
107                 a[1]+=tmp;tot-=tmp;
108
109                 tmp=Bit_Query(b1,m)-Bit_Query(b1,p[i].b)+Bit_Query(b2,p[i].a-1);
110                 a[3]+=tmp;tot-=tmp;
111
112                 a[2]+=tot;
113                 Bit_Add(b3,p[i].a,1);Bit_Add(b4,p[i].b,1);
114             }
115         }
116
117         for(int i=1;i<=n;i++){
118             if(p[i].a!=p[i].b)a[2]+=1;
119             else a[1]+=1;
120         }
121         printf("%lld %lld %lld %lld\n",a[1],a[2],a[3],a[4]);
122     }
123     return 0;
124 }