POJ 1113 Wall 凸包 裸
阿新 • • 發佈:2017-07-16
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題意:給出一個簡單幾何,問與其邊距離長為L的幾何圖形的周長。
思路:求一個幾何圖形的最小外接幾何,就是求凸包,距離為L相當於再多增加上一個圓的周長(因為只有四個角)。看了黑書使用graham算法極角序求凸包會有點小問題,最好用水平序比較好。或者用Melkman算法
/** @Date : 2017-07-13 14:17:05 * @FileName: POJ 1113 極角序求凸包 基礎凸包.cpp * @Platform: Windows * @Author : Lweleth ([email protected]) * @Link : https://github.com/ * @Version : $Id$ */ #include <stdio.h> #include <iostream> #include <string.h> #include <algorithm> #include <utility> #include <vector> #include <map> #include <set> #include <string> #include <stack> #include <queue> #include <math.h> //#include <bits/stdc++.h> #define LL long long #define PII pair<int ,int> #define MP(x, y) make_pair((x),(y)) #define fi first #define se second #define PB(x) push_back((x)) #define MMG(x) memset((x), -1,sizeof(x)) #define MMF(x) memset((x),0,sizeof(x)) #define MMI(x) memset((x), INF, sizeof(x)) using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5+20; const double eps = 1e-8; const double Pi = acos(-1.0); struct point { int x, y; point(){} point(int _x, int _y){x = _x, y = _y;} point operator -(const point &b) const { return point(x - b.x, y - b.y); } int operator *(const point &b) const { return x * b.x + y * b.y; } int operator ^(const point &b) const { return x * b.y - y * b.x; } }; double xmult(point p1, point p2, point p0) { return (p1 - p0) ^ (p2 - p0); } double distc(point a, point b) { return sqrt((double)((b - a) * (b - a))); } int n, l; point p[N]; stack<int>s; int cmp(point a, point b)//以p[0]基準 極角序排序 { int t = xmult(a, b, p[0]); if(t > 0) return 1; if(t == 0) return distc(a, p[0]) < distc(b, p[0]); if(t < 0) return 0; } void graham() { while(!s.empty()) s.pop(); for(int i = 0; i < min(n, 2); i++) s.push(i); int t = 1; for(int i = 2; i < n; i++) { while(s.size() > 1) { int p2 = s.top(); s.pop(); int p1 = s.top(); if(xmult(p[p1], p[p2], p[i]) > 0) { s.push(p2); break; } } s.push(i); } } int main() { while(~scanf("%d%d", &n, &l)) { int x, y; int mix, miy; mix = miy = INF; int pos = -1; for(int i = 0; i < n; i++) { scanf("%d%d", &x, &y); p[i] = point(x, y); if(miy > y || miy == y && mix > x)//註意選第一個點 是最左下方的 { mix = x, miy = y; pos = i; } } swap(p[pos], p[0]); sort(p + 1, p + n, cmp); graham(); double ans = l * Pi * 2.0000; int t = 0; while(!s.empty()) { ans += distc(p[t], p[s.top()]); t = s.top(); s.pop(); } printf("%d\n", (int)(ans + 0.500000)); } return 0; }
POJ 1113 Wall 凸包 裸