2
2 8 6
1 1
4 5
2 10 6
4 5
6 5

1
2解题思路：典型的区间完全覆盖问题。由于喷水装置是安置在横向中心线上并且圆具有对称性，故只需取高度的一半，然后将每个喷水装置能够覆盖的区间范围映射成在x轴的长度，然后按上面的方法贪心选线段即可。AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=10005;
 4 int t,n,k,cnt,pos,beg;double w,h,lb,xi,ri,maxv;pair<double,double> itv[maxn];bool flag;
 5 int main(){
 6     while(cin>>t){
 7         while(t--){
 8             cin>>n>>w>>h;h/=2.0;pos=cnt=0;//h取一半
 9             for(int i=0;i<n;++i){
10                 cin>>xi>>ri;
11                 if(ri<h)continue;//ri==h也要算
12                 itv[pos].first=xi-sqrt(ri*ri-h*h);
13                 itv[pos++].second=xi+sqrt(ri*ri-h*h);
14             }
15             sort(itv,itv+pos);lb=0;beg=0;flag=false;
16             if(itv[0].first>0){cout<<0<<endl;continue;}//按左端点排序只需查看最左边的端点是否满足条件即可，最右边的端点在下面有判断
17             while(lb<w){
18                 maxv=0;
19                 for(k=beg;k<pos&&itv[k].first<=lb;++k)//itv[k].first<=lb这样保证整个区间是连续的，即草坪都会被润湿
20                     maxv=max(maxv,itv[k].second);//找线段左端点在lb以内右端点能覆盖到的最远距离
21                 if(maxv>lb)cnt++,lb=maxv,beg=k;//如果有一条线段右端点比当前已覆盖的区间右端点lb还大，那么就更新已覆盖的右端点值，同时计数器加1
22                 else {flag=true;break;}//否则说明不能覆盖整个区间，直接退出，输出0
23             }
24             if(flag)cout<<0<<endl;
25             else cout<<cnt<<endl;
26         }
27     }
28     return 0;
29 }

2
5
2 3.2 4 4.5 6
10
1 2 3 1 2 1.2 3 1.1 1 2

2
5

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int t,n,cnt;double ans,ri,dt[605];
 4 int main(){
 5     while(cin>>t){
 6         while(t--){
 7             cin>>n;ans=0;cnt=0;
 8             for(int i=0;i<n;++i)cin>>ri,dt[i]=ri>1?sqrt(ri*ri-1):0;//这里可以设置为0，因为题目已经保证一定可以将草坪全部润湿
 9             sort(dt,dt+n);
10             for(int i=n-1;ans<=10.0&&i>=0;--i)cnt++,ans+=dt[i];//从后往前能选出最少数量的喷水装置，且一定能将草坪润湿
11             cout<<cnt<<endl;
12         }
13     }
14     return 0;
15 }

2
2
1 10
10 11
3
1 10
10 11
11 20

1
2解题思路：按结束时间早进行排序，然后贪心选择不相交区间即可。AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn=10005;
 5 int t,n,tmp,ans;pair<int,int> itv[maxn];
 6 int main(){
 7     while(cin>>t){
 8         while(t--){
 9             cin>>n;
10             for(int i=0;i<n;++i)cin>>itv[i].second>>itv[i].first;
11             sort(itv,itv+n);tmp=-1;ans=0;//按结束时间早进行排序
12             for(int i=0;i<n;++i)
13                 if(tmp<itv[i].second)ans++,tmp=itv[i].first;
14             cout<<ans<<endl;
15         }
16     }
17     return 0;
18 }

4
1 5
2 4
1 4
2 3
3
1 2
3 4
5 6
1
2 2

1
3
1解题思路：按左端点递增顺序排序，然后按上面的求解方法贪心选点即可。AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn=105;
 5 int n,tmp,ans;pair<int,int> itv[maxn];
 6 int main(){
 7     while(cin>>n){
 8         for(int i=0;i<n;++i)cin>>itv[i].first>>itv[i].second;
 9         sort(itv,itv+n);tmp=itv[0].second;ans=1;
10         for(int i=0;i<n;++i){
11             if(tmp<itv[i].first)ans++,tmp=itv[i].second;
12             else if(tmp>itv[i].second)tmp=itv[i].second;
13         }
14         cout<<ans<<endl;
15     }
16     return 0;
17 }

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 
We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 
 
Figure A Sample Input of Radar Installations

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 
The input is terminated by a line containing pair of zeros 

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

3 2
1 2
-3 1
2 11 2
0 20 0

Case 1: 2
Case 2: 1解题思路：典型的区间选点，将雷达能覆盖的范围映射为x轴上的线段长度，然后贪心区间选点即可。AC代码：

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cmath>
 4 using namespace std;
 5 const int maxn=1005;
 6 int n,d,ans,pos,cnt=1,x,y;double tmp;
 7 struct node{double l,r;}point[maxn];
 8 bool cmp(node a,node b){return a.l<b.l;}
 9 int main(){
10     while(cin>>n>>d&&(n|d)){//注意这里：n|d，表示n和d同时为0时，程序才退出
11         ans=1;pos=0;
12         for(int i=0;i<n;++i){
13             cin>>x>>y;
14             if(y>d){ans=-1;continue;}//根号下只能为非负数
15             point[pos].l=1.0*x-sqrt(1.0*d*d-y*y);//以每个岛屿为圆心，半径为d画圆，其与x轴最后只有两个交点
16             point[pos++].r=1.0*x+sqrt(1.0*d*d-y*y);
17         }
18         sort(point,point+pos,cmp);tmp=point[0].r;
19         for(int i=1;i<pos&&ans!=-1;++i){
20             if(tmp<point[i].l){ans++;tmp=point[i].r;}//如果已选线段与当前线段不相交，那么就设置一个新的雷达，然后更新tmp为其右端点值
21             else if(tmp>point[i].r)tmp=point[i].r;//可以覆盖掉下一条线段，但此时区间右端点缩短为下一条线段的右端点，说明覆盖的范围缩短了
22         }
23         cout<<"Case "<<cnt++<<": "<<ans<<endl;
24     }
25     return 0;
26 }