HDU 4343 贪心

D - Interval query
Time Limit: 1.5 Sec

Memory Limit: 256 MB

Description

This is a very simple question. There are N intervals in number axis, and M queries just like “QUERY(a,b)” indicate asking the maximum number of the disjoint intervals between (a,b) .

Input

There are several test cases. For each test case, the first line contains two integers N, M (0<N, M<=100000) as described above. In each following N lines, there are two integers indicate two endpoints of the i-th interval. Then come M lines and each line contains two integers indicating the endpoints of the i-th query. 
You can assume the left-endpoint is strictly less than the right-endpoint in each given interval and query and all these endpoints are between 0 and 1,000,000,000.

Output

For each query, you need to output the maximum number of the disjoint intervals in the asked interval in one line.

Sample Input

3 2
1 2
2 3
1 3
1 2
1 3

Sample Output

1
2

HINT

题意

              给你一些区间，现在有m个查询，求出每个查询的区间内的最大的不相交区间个数

题解：

             二分，比赛的时候这样写，我艹了一直超时，赛后看了qscqesze多加了一个判断，加上去结果过了，太弱了

代码：

   

///1085422276
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
typedef long long ll;
using namespace std;
#define inf 10000000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//***************************************************************

struct ss
{
    int x,y;
}a[100005],b[100005];
int lll;
int rrr;
bool cmp(ss a,ss b)
{
    return a.y<b.y||(a.y==b.y&&a.x>b.x);
}
int n,l,r;
int main()
{
    int m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0; i<n; i++)
        {
            a[i].x=read();
            a[i].y=read();
        }
        int tot=0;
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++)
        {
            int flag=0;
            for(int j=i+1;j<n;j++)
            {
                if(a[j].y>a[i].y)break;
                if(a[j].x<a[i].x)continue;
                flag=1;
                break;
            }
            if(!flag)b[tot++]=a[i];
        }
        n=tot;
        for(int i=1;i<=m;i++)
        {
            lll=read();
            rrr=read();
            int ans=0;
             l=0;r=n-1;
            int aa=-1;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(b[mid].y<=lll)
                {
                    aa=mid;
                    l=mid+1;
                }
                else r=mid-1;
            }
           for(int i=aa+1;i<n;i++)
           {
               if(b[i].y<=rrr)
               {
                   if(b[i].x>=lll)ans++,lll=b[i].y;
               }else break;
           }
           printf("%d
",ans);
        }

    }
    return 0;
}