【线段树】【4-6组队赛】Problem H

Problem Description

#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

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].

Output

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

Sample Input

4
1 1 1 2
1 1 1 2
1 1 1 1
1 1 1 1

Sample Output

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
",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();
    }
}