UVa 10535 - Shooter (最多区间覆盖的点 + 扫描)

题意

问一枪能最多能崩掉几堵墙

思路

求出墙和人的角度，一枪能崩掉这个角度里的墙，问题就转化成被覆盖最多的角度。

用左端点和右端点扫描。书上例题

因为对弧度不熟悉，所以转化为角度。。

如果相差大于180度，要分成[0, a][b, 360]。

代码

#include <stack>

#include <cstdio>

#include <list>

#include <set>

#include <iostream>

#include <string>

#include <vector>

#include <queue>

#include <functional>

#include <cstring>

#include <algorithm>

#include <cctype>

#include <string>

#include <map>

#include <cmath>

using namespace std;

#define LL long long

#define ULL unsigned long long

#define SZ(x) (int)x.size()

#define Lowbit(x) ((x) & (-x))

#define MP(a, b) make_pair(a, b)

#define MS(arr, num) memset(arr, num, sizeof(arr))

#define PB push_back

#define X first

#define Y second

#define ROP freopen("input.txt", "r", stdin);

#define MID(a, b) (a + ((b - a) >> 1))

#define LC rt << 1, l, mid

#define RC rt << 1|1, mid + 1, r

#define LRT rt << 1

#define RRT rt << 1|1

#define BitCount(x) __builtin_popcount(x)

#define BitCountll(x) __builtin_popcountll(x)

#define LeftPos(x) 32 - __builtin_clz(x) - 1

#define LeftPosll(x) 64 - __builtin_clzll(x) - 1

const double PI = acos(-1.0);

const int INF = 0x3f3f3f3f;

const double eps = 1e-8;

const int MAXN = 1e3 + 10;

const int MOD = 29;

const int dir[][2] = { {1, 0}, {0, 1} };

int cases = 0;

typedef pair<int, int> pii;

typedef pair<double, double> pdd;

struct LINE

{

pdd l, r;

}L[MAXN];

vector<pair<double, int> > range;

pdd pos;

int n;

void Convert()

{

range.clear();

for (int i = 0; i < n; i++)

{

LINE tmp = L[i];

double a = atan2(tmp.l.Y - pos.Y, tmp.l.X - pos.X) * 180 / PI, b = atan2(tmp.r.Y - pos.Y, tmp.r.X - pos.X) * 180 / PI;

if (a < 0) a += 360;

if (b < 0) b += 360;

if (a > b) swap(a, b);

if (b - a >= 180)

{

range.PB({0, 1}); range.PB({a, -1});

a = b; b = 360;

}

range.PB({a, 1}); range.PB({b, -1});

}

}

int cmp(pair<double, int> a, pair<double, int> b)

{

if (fabs(a.X - b.X) > eps) return a.X < b.X;

return a.Y > b.Y;

}

int main()

{

//ROP;

while (cin >> n, n)

{

    for (int i = 0; i < n; i++) cin >> L[i].l.X >> L[i].l.Y >> L[i].r.X >> L[i].r.Y;

cin >> pos.X >> pos.Y;

Convert();

sort(range.begin(), range.end(), cmp);

int tmp = 0, ans = -1;

for (int i = 0; i < SZ(range); i++)

{

tmp += range[i].Y;

ans = max(ans, tmp);

}

cout << ans << endl;

}

return 0;

}