Codeforces Gym 101164 H. Pub crawl (凸包)
2017-07-06 12:21
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题意
对于已知的 N 个二维坐标点,要求按照某种特定的连线方式将尽可能多的点连接(其中按照连线顺序任意相邻的 3 个点 i , j , k ,点 k 必须在有向线段 ij 的左侧。问最多可以连多少点,并给出连线顺序。解题思路
首先可以确定所有点都能够被按要求连接,同时一种有效连接方式为逆时针螺旋式收缩。通过类似凸包的策略将所有的点不断地压栈(即确认为凸包上的点)。唯一不同在于需对已经确认为当前凸包上的点标号作标记,并不断的对剩余的点继续做凸包。
代码
#include<bits/stdc++.h> using namespace std; const int N = 5000 + 10; int n; bool vis ; struct point { int x, y, idx; point() {}; point(int a, int b):x(a),y(b),idx(0){}; void input(int _idx) { scanf("%d %d",&x,&y); idx = _idx; } } a , stk ; bool cmp(point a, point b) { if(a.x == b.x) return a.y < b.y; return a.x < b.x; } double Cross(point &o, point &a, point &b) { double oax = a.x-o.x; double oay = a.y-o.y; double obx = b.x-o.x; double oby = b.y-o.y; return oax * oby - oay * obx; } void Convex() { int m = 0; while(m < n) { for(int i=0;i<n;i++) { if(vis[ a[i].idx ]) continue; while(m > 1 && Cross(stk[m-2], stk[m-1], a[i]) <= 0) { m--; vis[ stk[m].idx ] = 0; } stk[m++] = a[i]; vis[ a[i].idx ] = 1; } int mm = m; for(int i=n-1;i>=0;i--) { if(vis[ a[i].idx ]) continue; while(m > mm && Cross(stk[m-2], stk[m-1], a[i]) <= 0) { m--; vis[ stk[m].idx ] = 0; } stk[m++] = a[i]; vis[ a[i].idx ] = 1; } } } int main() { scanf("%d", &n); memset(vis, 0, sizeof(vis)); for(int i=0;i<n;i++) a[i].input(i+1); sort(a, a+n, cmp); Convex(); printf("%d\n", n); for(int i=0;i<n;i++) printf("%d ", stk[i].idx); }
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