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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\n",ans); } } return 0; }
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原文地址：http://www.cnblogs.com/zxhl/p/4746129.html
