SGU136 Erasing Edges

题目大意

给出一个N个顶点多边形所有边的中点，构造这个多边形，不存在则输出NO

算法思路

单独考虑X，对于某条边的中点有Xi=xi+xi+12，N个中点相当于N元一次方程组

• 当N为奇数时，方程组最终为x1∗2=Z的形式，回代得出所有解
• 当N为偶数时，方程组最终为0=Z的形式，无解则输出NO，否则回代任意值

Y的情况与X相同

时间复杂度: O(N)

代码

/** * Copyright © 2015 Authors. All rights reserved. *  * FileName: 136.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-06-14 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;const double eps = 1e-8;int sgn(double x) { return x < -eps? -1: x > eps; }const int maxn = 10000 + 5;int n;double mx[maxn], my[maxn], x[maxn], y[maxn];bool solve(){        if (n & 1) {                x[0] = 0;                rep(i,n) x[0] += (i & 1? -mx[i]: mx[i]) * 2;                x[0] /= 2;                rep(i,n-1) x[i+1] = mx[i] * 2 - x[i];                y[0] = 0;                rep(i,n) y[0] += (i & 1? -my[i]: my[i]) * 2;                y[0] /= 2;                rep(i,n-1) y[i+1] = my[i] * 2 - y[i];        } else {                x[0] = 0;                rep(i,n-1) x[i+1] = mx[i] * 2 - x[i];                if (sgn(x[n-1] + x[0] - mx[n-1] * 2)) return false;                y[0] = 0;                rep(i,n-1) y[i+1] = my[i] * 2 - y[i];                if (sgn(y[n-1] + y[0] - my[n-1] * 2)) return false;        }        puts("YES");        rep(i,n) printf("%.10f %.10f\n", x[i], y[i]);        return true;}int main(){        scanf("%d", &n);        rep(i,n) scanf("%lf%lf", &mx[i], &my[i]);        if (!solve()) puts("NO");        return 0;}

---