POJ 1328 Radar Installation(贪心)
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2022-03-26 14:06:21
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题目链接:http://poj.org/problem?id=1328
题意:平面直角坐标系上有n个点,求在x轴上找尽量少的点,以这些点为圆心画一个半径为d的圆,使得给定的点都在画出来的圆里。如果不能输出-1。
solution:围绕点画圆、将圆与x轴相交的区间都记录下来、在对这些区间利用贪心法、找出建雷达站的最少个数!
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
bool cmp(pair<double, double> &a, pair<double, double> &b)
{
if (a.second != b.second)return a.second < b.second;
return a.first > b.first;
}
int main()
{
int n, d, cas = 0;
while (cin >> n >> d){
if (!n && !d)return 0;
bool flag = true;
double x, y;
vector<pair<double, double> > vec;
for (int i = 0; i < n; ++i){
scanf("%lf%lf", &x, &y);
if (y > d)flag = false;
else vec.push_back(make_pair(x - sqrt(d * d - y * y), x + sqrt(d * d - y * y)));
}
sort(vec.begin(), vec.end(), cmp);
if (!flag)printf("Case %d: -1\n", ++cas);
else {
double end = vec[0].second, r;
int cnt = 1;
for (int i = 1; i < n; ++i)if (end < vec[i].first){
++cnt;
end = vec[i].second;
}
printf("Case %d: %d\n", ++cas, cnt);
}
}
}
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