【HDU

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
题意 给定n朵花的开放时间[s,e]，多次询问时间t 时有几朵花开。
分析，区间更新（加减），单点查询。明显的树状数组。
我用的是白书的离散化方法。
树状数组的常用操作
代码一

#include<bits/stdc++.h>using namespace std;typedef pair<int,int>pii;#define first fi#define second se#define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 2e5+10;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;struct BIT{    int n;    int c[MAXN<<1];    inline int lowbit(int x) { return x&(-x);}    void add(int x,int val){        for(int i=x;i<=n;i+=lowbit(i))            c[i]+=val;    }    int sum(int x){        int ans=0;        for(int i=x;i>0;i-=lowbit(i))             ans+=c[i];        return ans;    }}bit;struct Node{    int s,e;}Q[MAXN];int X[MAXN<<1],size; int ask[MAXN];int main(){    CLOSE();//  fread();//  fwrite();    int T;scanf("%d",&T);int ncase=0;    while(T--){        int n,m;scanf("%d%d",&n,&m);size=0;        n<<2;memset(bit.c,0,sizeof(bit.c));        for(int i=1;i<=n;i++){            scanf("%d%d",&Q[i].s,&Q[i].e);            X[size++]=Q[i].s; X[size++]=Q[i].e;        }        for(int i=1;i<=m;i++){            scanf("%d",&ask[i]);            X[size++]=ask[i];        }        bit.n=size+10;        sort(X,X+size);  size=unique(X,X+size)-X;        for(int i=1;i<=n;i++){            int l=lower_bound(X,X+size,Q[i].s)-X+1;            int r=lower_bound(X,X+size,Q[i].e)-X+1;        //  printf("l= %d r= %d\n",l,r);            bit.add(l,1);bit.add(r+1,-1);        }        printf("Case #%d:\n",++ncase);        for(int i=1;i<=m;i++){            int t=lower_bound(X,X+size,ask[i])-X+1;            printf("%d\n",bit.sum(t));        }    }    return 0;}

代码二，网上看大佬们的代码都是手写去重，手写二分查找，肯定有好处的 , 有哪位大佬路过指点一下 好在哪？Orz。
这里贴下代码：

#include<bits/stdc++.h>using namespace std;typedef pair<int,int>pii;#define first fi#define second se#define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 200008;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;struct BIT{    int n;    int c[MAXN<<1];    inline int lowbit(int x) { return x&(-x);}    void add(int x,int val){        for(int i=x;i<=n;i+=lowbit(i))            c[i]+=val;    }    int sum(int x){        int ans=0;        for(int i=x;i>0;i-=lowbit(i))             ans+=c[i];        return ans;    }}bit;struct Node{    int s,e;}Q[MAXN];int X[MAXN<<1],size,sz; int ask[MAXN];int Bin(int num,int R){      int l=0,r=R-1,mid;      while(l<=r){          mid=(l+r)>>1;         if(X[mid]==num) return mid;          if(X[mid]<num) l=mid+1;          else r=mid-1;      }  }  int main(){    CLOSE();//  fread();//  fwrite();    int T;scanf("%d",&T);int ncase=0;    while(T--){        int n,m;scanf("%d%d",&n,&m);        size=0,sz=1;          memset(bit.c,0,sizeof(bit.c));        for(int i=1;i<=n;i++){            scanf("%d%d",&Q[i].s,&Q[i].e);            X[size++]=Q[i].s; X[size++]=Q[i].e;        }        for(int i=1;i<=m;i++){            scanf("%d",&ask[i]);            X[size++]=ask[i];        }        // 离散         sort(X,X+size);          for(int i=1;i<size;i++)             if(X[i]!=X[i-1])  X[sz++]=X[i];        bit.n=sz+10;        for(int i=1;i<=n;i++){            int l=Bin(Q[i].s,sz)+1;            int r=Bin(Q[i].e,sz)+1;            bit.add(l,1);bit.add(r+1,-1);        }        printf("Case #%d:\n",++ncase);        for(int i=1;i<=m;i++){            int t=Bin(ask[i],sz)+1;            printf("%d\n",bit.sum(t));        }    }    return 0;}