D – Interval query
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88748#problem/D

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n个线段

m次询问，问你在ab区间内，最多放下几个不能相互重叠的线段

题解：

直接n*m暴力

首先，我们按照右端点排序，然后能放进去就放进去，这种策略肯定是最优的

然后就暴力吧……15s

有两个想一想好像并咩有什么用的剪枝：

预处理出来第一个能放进区间的线段

预处理出哪些线段永远都不可能选的

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define maxn 110000
using namespace std;

struct node
{
    int x,y;
};
bool cmp(node a,node b)
{
    if(a.y==b.y)
        return a.x<b.x;
    return a.y<b.y;
}
node a[maxn];
node b[maxn];
node qq[maxn];
int cc[maxn];
//node b[maxn];
int tot=0;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        tot=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(qq,0,sizeof(qq));
        memset(cc,0,sizeof(cc));
        for(int i=0;i<n;i++)
            scanf("%d %d",&a[i].x,&a[i].y);
        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];
        }
        int x,y,ans;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&qq[i].x,&qq[i].y);
            int l=0,r=tot-1;
            x = qq[i].x;
            y = qq[i].y;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(b[mid].y<=x)
                    l=mid+1;
                else
                    r=mid-1;
            }
            cc[i]=l;
        }
        for(int i=0;i<m;i++)
        {
            x = qq[i].x,y = qq[i].y;
            ans=0;
            int tmp=x;
            int l = cc[i];
            for(int j=l;j<tot;j++)
            {
                if(b[j].y<=y)
                {
                    if(b[j].x<tmp)
                        continue;
                    tmp = b[j].y;
                    ans++;
                }
                else
                    break;
            }
            printf("%d\n",ans);
        }
    }
}