题意：给一个凸多边形，求任选若干点形成的多边形的面积和。
思路：
- 按一定方向（顺时针或逆时针）对多边形的顶点进行编号，则多边形的面积计算公式为：f1 x f2 + f2 x f3 + ... fn-1 x fn + fn x f1，fi 表示从参考点到i的向量。
- 考虑fi x fj 在答案中出现的次数，则答案可以写成:fi x fj * 2n-(j-i+1) (i<j) 或 fi x fj * 2i-j-1(i>j)，直接枚举i和j是O(n2)的，显然不行。
- 由叉积的性质：∑(fi x P) = (∑fi) x P，可以预处理出fi*2i-1的前缀和以及fi/2i+1的前缀和，复杂度O(n)，从而只需枚举一维，另一维的和O(1)得到，总复杂度O(n)。
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#pragma comment(linker, "/STACK:10240000")
#include <bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
typedef long long ll;
typedef pair<int, int> pii;
namespace Debug {
void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<" ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
}
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
/* -------------------------------------------------------------------------------- */
const int maxn = 1e5 + 7;
const int mod = 1e9 + 7;
struct Point {
ll x, y;
void read() {
int x, y;
scanf("%d%d", &x, &y);
this->x = x;
this->y = y;
}
Point(int x, int y) {
this->x = x;
this->y = y;
}
Point() {}
Point operator + (const Point &that) const {
return Point((x + that.x) % mod, (y + that.y) % mod);
}
Point operator * (const int &m) const {
return Point(x * m % mod, y * m % mod);
}
};
Point p[maxn], g[maxn], h[maxn];
int two[maxn], twoinv[maxn];
int cross(Point &a, Point &b) {
return (a.x * b.y % mod - a.y * b.x % mod + mod) % mod;
}
int powermod(int a, int n, int mod) {
ll ans = 1, buf = a;
while (n) {
if (n & 1) ans = (ans * buf) % mod;
buf = buf * buf % mod;
n >>= 1;
}
return ans;
}
void init() {
ll inv = powermod(2, mod - 2, mod);
two[0] = twoinv[0] = 1;
for (int i = 1; i < maxn; i ++) {
two[i] = (two[i - 1] + two[i - 1]) % mod;
twoinv[i] = twoinv[i - 1] * inv % mod;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
int T, n;
cin >> T;
init();
while (T --) {
cin >> n;
for (int i = 1; i <= n; i ++) {
p[i].read();
g[i] = g[i - 1] + p[i] * two[i - 1];
h[i] = h[i - 1] + p[i] * twoinv[i + 1];
}
int ans = 0;
for (int i = 2; i <= n; i ++) {
Point buf = p[i] * two[n - i];
ans = (ans + cross(g[i - 1], buf)) % mod;
}
for (int i = 2; i <= n; i ++) {
Point buf = p[i] * two[i];
ans = (ans + cross(buf, h[i - 1])) % mod;
}
cout << ans << endl;
}
}
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时间： 2024-10-22 01:34:45