#include <iostream>

#include <algorithm>

using namespace std;



int n,a[110000],b[110000],c[110000],d[110000];

int main()

{

      while(cin>>n)

      {

            for(int i=0;i<n;++i)  cin>>a[i];

            for(int i=0;i<n;++i)  cin>>b[i];

            for(int i=0;i<n;++i)  cin>>c[i];

            for(int i=0;i<n;++i)  cin>>d[i];



            for(int i=0;i<n;++i)

                  if(a[i]>b[i])  swap(a[i],b[i]);

            for(int i=0;i<n;++i)

                  if(c[i]>d[i])  swap(c[i],d[i]);



            for(int i=0;i<n;++i)

            {

                  int ans=0;

                  for(int j=0;j<n;++j)

                        if(a[i]<=c[j]&&d[j]<=b[i])  ans++;

                  cout<<ans<<endl;

            }

      }

      return 0;

}





Even you are brave enough still, I really not recommend you to copy the code and just submit it.

Input contains several cases.

Each case begins with an integer n (0<n<=100000).

Then follow with 4 lines representing a[ ],b[ ],c[ ] and d[ ], all values are located in [1,100000].

For each case, just write out the answer as what the code do.

4

1 1 1 2

1 1 1 2

1 1 1 1

1 1 1 1

4

4

4

0

题目大意：

给你二组线段A,B 询问A组中每一个线段包含了多少B组线段

将A，B按左端点排序；

那么当B前面线段若不满足当前A了 ，就可以废弃不用了，即B.x<A[now].x。

所以未被删除的B线段都是满足了左端点，现在只需要求是否满足右端点即可。

用线段树来维护B的右端点。若B一条废弃不用，更新线段树即可。每次查询【0-A[now].y】有多少元素即可

nlogn的复杂度。

最后还要按序号排序回来，因为要按输入顺序输出，代码如下：

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <ctime>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <string>

#define oo 0x13131313

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

using namespace std;

const int maxn=100000+5;

struct node

{

    int x,y,num,ans;

}A[maxn],B[maxn];

int n;

int tree[maxn<<2];

bool cmp(node a,node b)

{

    return a.x<b.x;

}

bool cmp1(node a,node b)

{

    return a.num<b.num;

}

void pushup(int rt)

{

    tree[rt]=tree[rt<<1]+tree[rt<<1|1];

}

int build(int l,int r,int rt)

{

    if(l==r) {tree[rt]=0;return 0;}

    int m=(l+r)>>1;

    build(lson);

    build(rson);

    pushup(rt);

}

int updata(int p,int k,int l,int r,int rt)//单点更新

{

    int m;

    if(l==r) {tree[rt]+=k;return 0;}

    m=(l+r)>>1;

    if(p<=m) updata(p,k,lson);

    else updata(p,k,rson);

    pushup(rt);

}

int query(int L,int R,int l,int r,int rt)

{

    int temp=0,m;

    if(L<=l&&r<=R) return tree[rt];

    m=(l+r)>>1;

    if(L<=m) temp=temp+query(L,R,lson);

    if(R>m)  temp=temp+query(L,R,rson);

    return temp;

}

void input()

{

            for(int i=0;i<n;++i)  {scanf("%d",&A[i].x);A[i].num=i;}

            for(int i=0;i<n;++i)  scanf("%d",&A[i].y);

            for(int i=0;i<n;++i)  scanf("%d",&B[i].x);

            for(int i=0;i<n;++i)  scanf("%d",&B[i].y);

                for(int i=0;i<n;++i)

                  if(A[i].x>A[i].y)  swap(A[i].x,A[i].y);

                for(int i=0;i<n;++i)

                  if(B[i].x>B[i].y)  swap(B[i].x,B[i].y);

            sort(B,B+n,cmp);

            sort(A,A+n,cmp);

}

void solve()

{

     int tot=0;

     for(int i=0;i<n;i++)

     {

        updata(B[i].y,1,1,maxn-1,1);

     }

     for(int i=0;i<n;i++)

     {

        int ans=0;

        while(B[tot].x<A[i].x&&tot<n) {

                                    updata(B[tot].y,-1,1,maxn-1,1);

                                    tot++;

                               }

        ans=query(1,A[i].y,1,maxn-1,1);

        A[i].ans=ans;

     }

     sort(A,A+n,cmp1);

     for(int i=0;i<n;i++) printf("%d\n",A[i].ans);

}

void init()

{

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

}

int main()

{

   //  init();

	while(scanf("%d",&n)!=EOF)

    {

        input();

        build(1,maxn-1,1);

        solve();

    }

}