bzoj [Usaco2009 Hol]Cattle Bruisers 杀手游戏

Description

 

 

Input

 第1行输入N，R，BX，BY, BVX，BVY，之后N行每行输入四个整数Xi，Yi，VXi，VYi．

 

Output

 

    一个整数，表示在逃脱过程中，某一个时刻最多有这个数理的杀手可以射杀贝茜．

Sample Input

3 1 0 0 0 2
0 -3 0 4
1 2 -1 1
1 -2 2 -1

Sample Output

2

OUTPUT DETAILS:

At time 1.5, Bessie is at point (0, 3), and the three bruisers are
at points (0, 3), (-0.5, 3.5), and (4, -3.5). The first two cattle
bruisers are within 1 unit of Bessie, while the third will never
be within 1 unit of Bessie, so 2 is the most achievable.

 

思路:  我们可以计算出每个杀手可以射击的一个时间范围，然后做扫描线。时间复杂度$O(nlogn)$

#include<bits/stdc++.h>
using namespace std;
int const N = 50000 + 3;
double const eps = 1e-8;
int n, r, m, in[N], out[N], cnt;
double bx, by, bvx, bvy, x[N], y[N], vx[N], vy[N], t1[N], t2[N], t[N << 2];
void calc(int k) {
    double tx = x[k] - bx;
    double ty = y[k] - by;
    double v1 = vx[k] - bvx;
    double v2 = vy[k] - bvy;
    double a = v1 * v1 + v2 * v2;
    double b = 2 * v1 * tx + 2 * v2 * ty;
    double c = tx * tx + ty * ty - 1.0*r * r;
    double d = b * b - 4 * a * c;
    if(fabs(a) < eps ) {
        if( c<0) { // 这个我始终觉得很奇怪，为什么c小于0就无限解了。
            m++;
            t1[m] = 0;
            t2[m] = 1e10;
        }
        return ;
    }
    if(d<0) return ;
    t1[k] = (-b - sqrt(d)) / (2 * a);
    t2[k] = (-b + sqrt(d)) / (2 * a);
    if(t2[k] <0)
        return ;
    t1[k] = max(0.0, t1[k]);
    m++;
    t1[m] = t1[k];
    t2[m] = t2[k];
}
int main() {
    scanf("%d%d%lf%lf%lf%lf", &n, &r, &bx, &by, &bvx, &bvy);
    for(int i = 1; i <= n; i++) {
        scanf("%lf%lf%lf%lf", &x[i], &y[i], &vx[i], &vy[i]);
        calc(i);
    }
    for(int i = 1; i <= m; i++) {
        ++cnt;
        t[cnt] = t1[i];
        ++cnt;
        t[cnt] = t2[i];
    }
    sort(t + 1, t + cnt + 1);
    int c = unique(t, t + cnt + 1) - t - 1;
    for(int i = 1; i <= m; i++) {
        int a = lower_bound(t + 1, t + c + 1, t1[i]) - t;
        int b = lower_bound(t + 1, t + c + 1, t2[i]) - t;
        in[a]++;
        out[b]++;
    }
    int ans = 0, now = 0;
    for(int i = 1; i <= c; i++) {
        now += in[i];
        ans = max(ans, now);
        now -= out[i];
    }
    printf("%d\n", ans);
    return 0;
}