hdoj 4325 Flowers【线段树+离散化】
Flowers
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2616 Accepted Submission(s):
1287
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 Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, means the time of i-th query.
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 Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, 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 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
题意：给出花的种类n以及m个要查询的时间，接下来n行是每种花的开放区间（其中一些花的开放区间有重合的部分），接下来m个数非别表示要查询的点（查询在这个时间点，有多少种花开放）
题解：直接线段树建图的话由于数据达到了10的9次方，数组无法开这么大，这就需要用到离散化
离散化：在百科上看到一句很好的话：“离散化就是把连续的量，变成离散的量即变成一个一个的值”，例如区间（1,100）由于这是一个实数区间，其中间的值有无数个，如果我们能把它变为1到100内的整数，这样这些数就变成了有限个，即离散了； ，我们将每个区间的端点都存到一个数组中，然后将这些端点，按照从小到大排列（之后要去除这个数组中重复的点），并建立为与其下标的映射，然后用其下标建树，因为我们只是使用了需要的空间，并没有在整个空间上建树，这样就大大节省了空间和时间，如题中第二个例子，我们将所有数据按照从小到大排列后为1 2 3 4 6 8 分别对应下标1 2 3 4 5 6，此题数据小我们看不出明显的差别，但是如果数据中有区间（1000,10000），那差别马上就出来了，比如我们把题中的区间（4,8）换做（1000,10000）那么如果采用离散化思想，我们还是先排列大小 1 2 3 6 1000 10000对应下标1 2 3 4 5 6，我们只需建一棵根为6的树即可，如果不用离散化，我们就需要建造根为10000的树，大大浪费了空间，
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
#define MAX 100100
using namespace std;
int s[MAX],e[MAX],q[MAX];
int rec[MAX];//记录所有值排序后的下标
int add[MAX<<2];
int sum[MAX<<2];
int le[MAX],ri[MAX];
void pushup(int o)
{
sum[o]=sum[o<<1]+sum[o<<1|1];
}
void pushdown(int o,int m)
{
if(add[o])
{
add[o<<1]+=add[o];
add[o<<1|1]+=add[o];
sum[o<<1]+=add[o]*(m-(m>>1));
sum[o<<1|1]+=add[o]*(m>>1);
add[o]=0;
}
}
void gettree(int o,int l,int r)
{
add[o]=0;
if(l==r)
{
sum[o]=0;
return ;
}
int mid=(l+r)>>1;
gettree(o<<1,l,mid);
gettree(o<<1|1,mid+1,r);
pushup(o);
}
void update(int o,int l,int r,int L,int R,int val)
{
if(L<=l&&R>=r)
{
add[o]+=val;
sum[o]+=val*(r-l+1);
return ;
}
pushdown(o,r-l+1);
int mid=(l+r)>>1;
if(L<=mid)
update(o<<1,l,mid,L,R,val);
if(R>mid)
update(o<<1|1,mid+1,r,L,R,val);
pushup(o);
}
int find(int o,int l,int r,int pos)
{
if(l==r)
{
return sum[o];
}
pushdown(o,r-l+1);
int ans=0;
int mid=(l+r)>>1;
if(pos<=mid)
ans=find(o<<1,l,mid,pos);
else
ans=find(o<<1|1,mid+1,r,pos);
return ans;
}
int query(int l,int r,int pos)//查找输入当前值，在树中对应的位置
{
while(r>=l)
{
int mid=(l+r)>>1;
if(rec[mid]==pos)
return mid;
else if(rec[mid]>pos)
r=mid-1;
else
l=mid+1;
}
return -1;
}
int main()
{
int t,n,m,k,i;
scanf("%d",&t);
k=1;
int maxx;
while(t--)
{
scanf("%d%d",&n,&m);
int p=1;
for(i=0;i<n;i++)
{
scanf("%d%d",&s[i],&e[i]);
rec[p++]=s[i];
rec[p++]=e[i];
}
for(i=0;i<m;i++)
{
scanf("%d",&q[i]);
rec[p++]=q[i];
}
sort(rec+1,rec+p);//
int R=2;
for(i=2;i<p;i++)//去除数组中重复的点
{
if(rec[i]!=rec[i-1])
rec[R++]=rec[i];
}
sort(rec+1,rec+R);
gettree(1,1,R-1);//对下标建树
for(int i=0;i<n;i++)
{
int x=query(1,R-1,s[i]);
int y=query(1,R-1,e[i]);
update(1,1,R-1,x,y,1);
}
printf("Case #%d:\n",k++);
for(i=0;i<m;i++)
{
int x=query(1,R-1,q[i]);
printf("%d\n",find(1,1,R-1,x));
}
}
}
