贪心-poj1328

Posted 2020-08-23

http://poj.org/problem?id=1328

求出每个岛屿点在x轴上对应的区间(雷达在此区间内能探测到该岛屿).

将区间按照区间左端升序,然后贪心求出最小的相交区间个数.

坑:y^2爆int,爆long long.用double可过

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
struct haha{
    double xz,xy;
}a[1010];
bool operator <(haha aa,haha bb){
        return aa.xy<bb.xy;
}
int main(){
    int T,n,d;
    double x,y;
    double tmp;
    int flag=0;
    while(cin>>n>>d && n){
        T++;
        flag=0;
        if(d<0)flag=-1;
        for(int i=0;i<n;i++){
            cin>>x>>y;
            if(flag==0){
                if(y>d || y<0){
                    flag=-1;
                }else{
                    tmp=sqrt(d*d-y*y);
                    a[i].xz=(double)x-tmp;
                    a[i].xy=(double)x+tmp;
                }
            }
        }
        sort(a,a+n);
        double xyy=a[0].xy;
        if(flag!=-1){
            flag=1;
            for(int i=1;i<n;i++){
                if(a[i].xz>xyy){
                    xyy=a[i].xy;
                    flag++;
                }
            }
        }
        printf("Case %d: %d\\n",T,flag);
    }
    return 0;
}

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