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Description
Input
Output
Sample Input
3 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1
大概题义就是要在x轴上放圆心,使得放最少的圆可以把所有点覆盖,输出圆的个数
#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define Maxn 10010
struct Node{
double a,b;
}A[Maxn];
bool cmp(Node a,Node b){
if(a.a != b.a){
return a.a < b.a;
}else{
return a.b < b.b;
}
}
double wenbao(double R,double high){
return sqrt( pow(R,2.0) - pow(high,2.0) );
}
int main(){
int N;
double R;
double a,b;
int count = 1;
while(cin >> N >> R,N+R){
bool flag = true;
for(int i = 0; i < N; i++){
// cin >> a >> b;
scanf("%lf%lf",&a,&b);//用scanf,否则会超时
if(fabs(b) > R){
flag = false;
}else{
double temp = wenbao(R,b);
A[i].a = a - temp;
A[i].b = a + temp;
}
}
sort(A,A+N,cmp);
Node p = A[0];
int cnt = 1;
for(int i = 1; i < N; i++){
if(A[i].a > p.b){
cnt++;
p = A[i];
}else if(A[i].b < p.b){//请仔细思考
p = A[i];
}
}
if(flag){
printf("Case %d: %d\n",count++,cnt);
}else{
printf("Case %d: -1\n",count++);
}
}
}
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原文地址:http://www.cnblogs.com/yakoazz/p/5839247.html