UVA - 10382 Watering Grass（几何）

Posted 2020-10-04 SomnusMistletoe

题意：有一个矩形，n个圆。已知矩形的长宽和圆的半径，问最少需多少个圆将矩形完全覆盖。

分析：

1、首先求圆与矩形的长的交点，若无交点，则一定不能对用最少的圆覆盖矩形有贡献。

2、如果两个圆与矩形相交所得的线段重合，那这两个圆一定能把矩形在两线段并集的那部分所覆盖。问题转化为用圆与矩形相交所得的线段覆盖矩形的长。

3、按线段左端点排序，对于某个已选择的线段a，求它后面满足b.L <= a.R的线段b的b.R的最大值，依次类推。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
struct Node{
    double pos, r;
    double L, R;
    bool operator < (const Node& rhs)const{
        return L < rhs.L;
    }
}num[MAXN];

int main(){
    int n;
    double l, w;
    while(scanf("%d%lf%lf", &n, &l, &w) == 3){
        double pos, r;
        int cnt = 0;
        for(int i = 0; i < n; ++i){
            scanf("%lf%lf", &pos, &r);
            double tmpl = r * r - w * w / 4.0;
            if(dcmp(tmpl, 0.0) <= 0) continue;
            ++cnt;
            num[cnt].pos = pos;
            num[cnt].r = r;
            num[cnt].L = pos - sqrt(tmpl);
            num[cnt].R = pos + sqrt(tmpl);
        }
        sort(num + 1, num + cnt + 1);
        int ans = 0;
        double st = 0.0;
        bool ok = false;
        for(int i = 1; i <= cnt; ++i){
            if(dcmp(st, l) >= 0){
                ok = true;
                break;
            }
            ++ans;
            if(num[i].L > st){
                ok = false;
                break;
            }
            double ma = -1.0;
            int j;
            for(j = i; j <= cnt; ++j){
                if(dcmp(num[j].L, st) <= 0){
                    ma = max(ma, num[j].R);
                }
                else break;
            }
            st = ma;
            i = j - 1;
        }
        if(dcmp(st, l) >= 0){
            ok = true;
        }
        if(ok){
            printf("%d\n", ans);
        }
        else{
            printf("-1\n");
        }
    }
    return 0;
}