【线段树】 hdu4325 Flowers

Flowers

http://acm.hdu.edu.cn/showproblem.php?pid=4325

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

21 15 1042 31 44 8146

 

Sample Output

Case #1:0Case #2:121

题意：给出花开放和凋谢的时间点，询问在一个时间点上有多少花同时开放。

题解：成段更新，单点查询的线段树，就是花开谢时间要hash处理，不然必然超内存。就是这个hash处理个人觉得我做的有点麻烦了。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MAX 100005struct node{    int s,t;}num[MAX];int bloom[MAX<<2];int add[MAX<<2];void build(int l,int r,int idx){    add[idx]=0;    if(l==r) return;    int mid=(l+r)>>1;    build(l,mid,idx<<1);    build(mid+1,r,idx<<1|1);}int binarySearch(int x,int n){    int l=0,r=n-1,mid;    for(;l<=r;)    {        mid=(l+r)>>1;        if(x<bloom[mid])           r=mid-1;        else if(x==bloom[mid])           return mid;        else           l=mid+1;    }    return l;}void push_down(int l,int r,int idx){    if(add[idx])    {        add[idx<<1]+=add[idx];        add[idx<<1|1]+=add[idx];        add[idx]=0;    }}void update(int a,int b,int l,int r,int idx){    if(a<=l&&r<=b)    {        add[idx]++;        return;    }    push_down(l,r,idx);    int mid=(l+r)>>1;    if(a<=mid) update(a,b,l,mid,idx<<1);    if(b>mid)  update(a,b,mid+1,r,idx<<1|1);}int query(int x,int l,int r,int idx){    if(l==x&&x==r)       return add[idx];    push_down(l,r,idx);    int mid=(l+r)>>1;    if(x<=mid) return query(x,l,mid,idx<<1);    else       return query(x,mid+1,r,idx<<1|1);}int main(){    int ti,n,m,cnt,ans;    scanf("%d",&ti);    for(int cas=1;cas<=ti;++cas)    {        cnt=1;        printf("Case #%d:\n",cas);        scanf("%d%d",&n,&m);        for(int i=0;i<n;++i)        {            scanf("%d%d",&num[i].s,&num[i].t);            bloom[cnt++]=num[i].s;            bloom[cnt++]=num[i].t;        }        //以下就是hash处理，就是把时间点的-1和+1都补上        bloom[0]=0;        sort(bloom,bloom+cnt);        ans=1;        for(int i=1;i<cnt;++i)            if(bloom[i]!=bloom[i-1]) bloom[ans++]=bloom[i];        for(int i=ans-1;i>0;--i)            if(bloom[i]!=bloom[i-1]+1) bloom[ans++]=bloom[i-1]+1;        sort(bloom,bloom+ans);        for(int i=ans;i>0;--i)            if(bloom[i]!=bloom[i-1]+1) bloom[ans++]=bloom[i]-1;        sort(bloom,bloom+ans);        //hash处理完毕        build(0,ans,1);        for(int i=0;i<n;++i)            update(binarySearch(num[i].s,ans),binarySearch(num[i].t,ans),0,ans,1);        for(;m--;)        {            scanf("%d",&cnt);            int pos=binarySearch(cnt,ans);            printf("%d\n",query(pos,0,ans,1));        }    }    return 0;}