链接：
http://poj.org/problem?id=1265
题意：
给你一个多边形，求它的面积，内部格点数目，边上格点数目
题解：
pick公式：
给定顶点坐标均是整数点的简单多边形，有
面积=内部格点数目+边上格点数目/2+1
边界上的格点数：
把每条边当做左开右闭的区间以避免重复，一条左开右闭的线段(x1,y1)->(x2,y2)上的格点数为：
gcd(x2-x1,y2-y1)。
代码：
1 #include <map>
2 #include <set>
3 #include <cmath>
4 #include <queue>
5 #include <stack>
6 #include <cstdio>
7 #include <string>
8 #include <vector>
9 #include <cstdlib>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <functional>
15 using namespace std;
16 #define rep(i,a,n) for (int i=a;i<n;i++)
17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
18 #define all(x) (x).begin(),(x).end()
19 #define pb push_back
20 #define mp make_pair
21 #define lson l,m,rt<<1
22 #define rson m+1,r,rt<<1|1
23 typedef long long ll;
24 typedef vector<int> VI;
25 typedef pair<int, int> PII;
26 const ll MOD = 1e9 + 7;
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2e4 + 7;
29 // head
30
31 const double eps = 1e-8;
32 int cmp(double x) {
33 if (fabs(x) < eps) return 0;
34 if (x > 0) return 1;
35 return -1;
36 }
37
38 const double pi = acos(-1);
39 inline double sqr(double x) {
40 return x*x;
41 }
42 struct point {
43 double x, y;
44 point() {}
45 point(double a, double b) :x(a), y(b) {}
46 void input() {
47 scanf("%lf%lf", &x, &y);
48 }
49 friend point operator+(const point &a, const point &b) {
50 return point(a.x + b.x, a.y + b.y);
51 }
52 friend point operator-(const point &a, const point &b) {
53 return point(a.x - b.x, a.y - b.y);
54 }
55 friend point operator*(const double &a, const point &b) {
56 return point(a*b.x, a*b.y);
57 }
58 friend point operator/(const point &a, const double &b) {
59 return point(a.x / b, a.y / b);
60 }
61 double norm() {
62 return sqrt(sqr(x) + sqr(y));
63 }
64 };
65 double det(point a, point b) {
66 return a.x*b.y - a.y*b.x;
67 }
68 double dot(point a, point b) {
69 return a.x*b.x + a.y*b.y;
70 }
71 double dist(point a, point b) {
72 return (a - b).norm();
73 }
74
75 struct line {
76 point a, b;
77 line() {}
78 line(point x, point y) :a(x), b(y) {}
79 };
80 double dis_point_segment(point p, point s, point t) {
81 if (cmp(dot(p - s, t - s)) < 0) return (p - s).norm();
82 if (cmp(dot(p - t, s - t)) < 0) return (p - t).norm();
83 return fabs(det(s - p, t - p) / dist(s, t));
84 }
85 bool point_on_segment(point p, point s, point t) {
86 return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0;
87 }
88 bool parallel(line a, line b) {
89 return !cmp(det(a.a - a.b, b.a - b.b));
90 }
91 bool line_make_point(line a, line b,point &res) {
92 if (parallel(a, b)) return false;
93 double s1 = det(a.a - b.a, b.b - b.a);
94 double s2 = det(a.b - b.a, b.b - b.a);
95 res = (s1*a.b - s2*a.a) / (s1 - s2);
96 return true;
97 }
98
99 struct polygon {
100 int n;
101 point a[MAXN];
102 double area() {
103 double sum = 0;
104 a[n] = a[0];
105 rep(i, 0, n) sum += det(a[i], a[i + 1]);
106 return sum / 2;
107 }
108 point MassCenter() {
109 point ans = point(0, 0);
110 if (cmp(area()) == 0) return ans;
111 a[n] = a[0];
112 rep(i, 0, n) ans = ans + det(a[i + 1], a[i])*(a[i] + a[i + 1]);
113 return ans / area() / 6;
114 }
115 int gcd(int a, int b) {
116 return b == 0 ? a : gcd(b, a%b);
117 }
118 int Border_Int_Point_Num() {
119 int num = 0;
120 a[n] = a[0];
121 rep(i, 0, n) num += gcd(abs(int(a[i + 1].x - a[i].x)),
122 abs(int(a[i + 1].y - a[i].y)));
123 return num;
124 }
125 int Inside_Int_Point_Num() {
126 return int(area()) + 1 - Border_Int_Point_Num() / 2;
127 }
128 };
129
130 int n;
131 polygon p;
132
133 int main() {
134 int T;
135 cin >> T;
136 rep(cas, 1, T + 1) {
137 cin >> n;
138 p.n = n;
139 rep(i, 0, n) scanf("%lf%lf", &p.a[i].x, &p.a[i].y);
140 rep(i, 1, n) p.a[i] = p.a[i] + p.a[i - 1];
141 int a = p.Inside_Int_Point_Num();
142 int b = p.Border_Int_Point_Num();
143 double c = p.area();
144 printf("Scenario #%d:\n%d %d %.1f\n\n", cas, a, b, c);
145 }
146 return 0;
147 }
时间： 2024-10-13 22:23:26