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主要思路:求出蚊子到达球的时间区间(用方程得解),对区间做一个贪心的选择,选择尽可能多的区间有交集的区间段(结构体排序即可),然后计数.
#include <cstdio> #include <cmath> #include <iostream> #include <algorithm> using namespace std; #define ll long long #define maxn 100025 int n, m, x,y; ll r; int t; struct point { ll x, y, z,dx,dy,dz; point(ll x = 0, ll y = 0, ll z = 0, ll dx = 0, ll dy = 0, ll dz = 0) :x(x), y(y), z(z) {} } a[maxn]; ll dis(point p1,point p2) { return (p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y) + (p1.z - p2.z)*(p1.z - p2.z); } int b[maxn]; double c[maxn]; struct node { double lf, ri; } lr[maxn]; bool cmp(node a, node b) { return a.lf < b.lf; } int main() { int cas = 1; scanf("%d", &t); while (t--) { scanf("%d%ld", &n, &r); int k = 0; point a; for (int i = 0; i < n; i++) { scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &a.x, &a.y, &a.z, &a.dx, &a.dy, &a.dz); ll bb = 2 * (a.x*a.dx + a.y*a.dy + a.z*a.dz); ll aa = a.dx*a.dx + a.dy*a.dy + a.dz*a.dz; ll cc = a.x*a.x + a.y*a.y + a.z*a.z - r*r; if (bb*bb - 4 * aa*cc >= 0) { if (dis(a, point(0, 0, 0))>r*r&&a.x*a.dx + a.y*a.dy +a.z*a.dz >= 0)continue; double m1 = max(0.0, (-bb - sqrt((double)bb*bb - 4 * (double)aa*cc)) / 2 / aa); double m2 = max(0.0, (-bb + sqrt((double)bb*bb - 4 * (double)aa*cc)) / 2 / aa); lr[k].lf = min(m1, m2), lr[k++].ri = max(m1, m2); } } sort(lr, lr + k,cmp); c[k - 1] = lr[k-1].ri; for (int i = k - 2; i >= 0; i--)c[i] = min(c[i + 1], lr[i].ri); printf("Case %d: ", cas++); double temp; m = 0; for (int i = 0; i < k;) { temp = c[i]; m++; while (i<k&&lr[i].lf <= temp)i++; } printf("%d %d\n", k, m); } }
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原文地址：http://www.cnblogs.com/jifahu/p/5448816.html
