uva 10535 - Shooter(几何+最大区间覆盖）

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题目链接：uva 10535 - Shooter

题目大意：给出若干堵墙（线段），然后给出人的位置，问说这个人最多开枪射穿几堵墙。

解题思路：最大区间覆盖问题的变形，首先把每堵墙转化成一个角度，然后计算角度区间的最大覆盖，注意：超过pi的要分成两个区间。

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;const double eps = 1e-9;const double pi = acos(-1.0);const int N = 1005;struct state {double t;int k;}s[N*2];struct point {double x1, x2, y1, y2;}p[N];int n, m;double x, y;bool cmp(const state& a, const state& b) {if (fabs(a.t - b.t) > eps) return a.t < b.t;return a.k > b.k;}void init() {m = 0;for (int i = 0; i < n; i++) scanf("%lf%lf%lf%lf", &p[i].x1, &p[i].y1, &p[i].x2, &p[i].y2);scanf("%lf%lf", &x, &y);for (int i = 0; i < n; i++) {double l = atan2(p[i].y1 - y, p[i].x1 - x);double r = atan2(p[i].y2 - y, p[i].x2 - x);if (l > r) swap(r, l);if (r - l >= pi) {s[m].t = -pi; s[m++].k = 1;s[m].t = l;   s[m++].k = -1;l = r; r = pi;}s[m].t = l; s[m++].k = 1;s[m].t = r; s[m++].k = -1;}sort(s, s + m, cmp);}int solve() {int ans = 0, c = 0;for (int i = 0; i < m; i++) {c += s[i].k;ans = max(ans, c);}return ans;}int main() {while (scanf("%d", &n) == 1 && n) {init();printf("%d\n", solve());}return 0;}

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