hdu 4325 Flowers(线段树+哈希散列)

Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1076    Accepted Submission(s): 535

Problem Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times.
In the next N lines, each line contains two integer S _i and T _i (1 <= S _i <= T _i <= 10^9), means i-th flower will be blooming at time [S _i, T _i].
In the next M lines, each line contains an integer T _i, means the time of i-th query.

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

Sample Input

  

   2
1 1
5 10
4
2 3
1 4
4 8
1
4
6
  

Sample Output

  

   Case #1:
0
Case #2:
1
2
1
  

Author

Cloud

Source

2012 Multi-University Training Contest 3

Recommend

zhoujiaqi2010

 

线段树使用hash，这次编程犯了个极其低级的错误，
f[j]写成了f[i]
导致调了一下午，还有build()里面的初始化。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100300
using namespace std;
class node
{
  public:
  int l,r,lazy,val;
}root[N<<3];
class A
{
  public:
  int l,r;
}q[N];
int hash[N<<3],len;
int f[N];
int ss(int pos)
{
  int l=0,r=len,mid=len/2;
  while(l<=r)
  {
    if(hash[mid]==pos)return mid;
    else if(hash[mid]>pos)
    {
      r=mid-1;mid=(l+r)/2;
    }
    else {l=mid+1;mid=(l+r)/2;}
  }
}
void build(int t,int l,int r)
{
  root[t].l=l;root[t].r=r;
  root[t].lazy=0;root[t].val=0;//就是这原先没加上WA
  if(root[t].l==root[t].r){root[t].lazy=0;root[t].val=0;return;}
  build(t<<1,l,(l+r)/2);build(t<<1|1,(l+r)/2+1,r);
}
void update(int t,int l,int r)
{
  if(root[t].l==l&&root[t].r==r){root[t].lazy++;root[t].val++;}
  else
  {
    if(root[t].lazy)
    {
      root[t<<1].lazy+=root[t].lazy;
      root[t<<1].val+=root[t].lazy;
      root[t<<1|1].lazy+=root[t].lazy;
      root[t<<1|1].val+=root[t].lazy;
     // root[t<<1].val=root[t].val;root[t<<1|1].val=root[t].val;
      root[t].lazy=0;//root[t].val=0;
    }
    if(root[t<<1].r>=r)update(t<<1,l,r);
    else if(root[t<<1|1].l<=l)
    {
      update(t<<1|1,l,r);
    }
    else
    {update(t<<1,l,root[t<<1].r);
     update(t<<1|1,root[t<<1|1].l,r);
    }
  }
}
int query(int t,int pos)
{
  if(root[t].l==root[t].r&&root[t].l==pos)
  return root[t].val;
  if(root[t].lazy)
  {
    root[t<<1].val+=root[t].lazy;root[t<<1|1].val+=root[t].lazy;
    root[t<<1].lazy+=root[t].lazy;root[t<<1|1].lazy+=root[t].lazy;
    root[t].lazy=0;
  }
  if(root[t<<1].r>=pos)
  query(t<<1,pos);
  else query(t<<1|1,pos);
}
int main()
{
  int cas;
  int m,n;
  scanf("%d",&cas);
  for(int i=1;i<=cas;i++)
  {  len=0;
      scanf("%d%d",&n,&m);
      for(int j=0;j<n;j++)
      {scanf("%d%d",&q[j].l,&q[j].r);hash[len++]=q[j].l;hash[len++]=q[j].r;}
      for(int j=0;j<m;j++)
      {
        scanf("%d",&f[j]);hash[len++]=f[j];
      }
      sort(hash,hash+len);
      int k=0;
      for(int j=1;j<len;j++)
      if(hash[k]!=hash[j])
      hash[++k]=hash[j];
      len=k;
      build(1,0,len);
      for(int j=0;j<n;j++)
      update(1,ss(q[j].l),ss(q[j].r));
      printf("Case #%d:\n",i);
      for(int j=0;j<m;j++)
      printf("%d\n",query(1,ss(f[j])));
  }
}