hdu6325 Interstellar Travel 凸包变形
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题目大意:
给出n个平面坐标,保证第一个点和第n个点y值为0,其余点的x坐标都在中间,要从 i 点走到 j 点的要求是 i 点的横坐标严格小于 j 的横坐标,并且消耗的能量是(xi * yj - xj * yi),要求消耗的能量最小(能量可以为负),并且字典序要求最小。
思路:
消耗能量的式子就是两个坐标的叉积,叉积的几何意义就是两个向量对应的平行四边形的面积,但是这个面积会有正负,如果向量 j 在向量 i 的顺时针方向,则这个面积是负的,如果我希望能量最小,那么就尽可能使向量是顺时针方向的,由此发现其实就得到了一个凸包,而且是一个上凸包。经过凸包上的点都符合能量的要求,但是由于要求字典序最小,所以如果凸包上的某一条边上有很多点,那么就需要判断一下这些点的id,如果线段中间的点的id比较小,那么这些点选上了之后,字典序肯定会变小,所以在做凸包的时候就要对凸包算法加一点点变形。
#include<cstdio> #include<algorithm> #include<iostream> #include<vector> #include<map> #include<set> #include<cstring> #include<queue> #include<stack> #define CLR(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) typedef long long ll; using namespace std; const int inf = 0x3f3f3f3f; const int maxn = 200010; struct dian { double x, y; int id; dian (){} dian(double x, double y, int id) :x(x), y(y), id(id){} }a[maxn],ch[maxn]; bool cmp(dian &a, dian &b) { if (a.x - b.x) return a.x < b.x; if (a.y - b.y) return a.y < b.y; return a.id < b.id; } typedef dian Vector; Vector operator -(Vector a, Vector b) { return Vector ( a.x - b.x, a.y - b.y,0 ); } bool operator == (Vector a,Vector b){ return ((a.x==b.x) && (a.y==b.y)); } double cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x; } int andrew(dian *p, int n, dian *ch) { int m = 0; sort(p, p + n, cmp); for (int i = 0; i < n; i++) { if(i>0&&p[i]==ch[m-1])continue; while (m > 1 && cross(ch[m - 2] - ch[m - 1], p[i] - ch[m - 1] )<= 0) { if(cross(ch[m - 2] - ch[m - 1], p[i] - ch[m - 1] )< 0) m--; else if(ch[m-1].id>p[i].id){ m--; }else break; } ch[m++] = p[i]; } return m; } bool vis[maxn]; int main() { int n,T; cin >> T; while (T--) { CLR(vis, inf); scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%lf%lf", &a[i].x, &a[i].y); a[i].id = i+1; } int m = andrew(a, n, ch); for (int i = 0; i<=m-1; i++) { printf("%d", ch[i].id); if (i < m-1)printf(" "); else printf(" "); } } } /* 1 7 0 0 9 0 3 6 1 2 3 6 2 4 10 0 */
Problem G. Interstellar Travel
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2098 Accepted Submission(s): 544
Little Q knows the position of n planets in space, labeled by 1 to n. To his surprise, these planets are all coplanar. So to simplify, Little Q put these n planets on a plane coordinate system, and calculated the coordinate of each planet (xi,yi).
Little Q plans to start his journey at the 1-th planet, and end at the n-th planet. When he is at the i-th planet, he can next fly to the j-th planet only if xi<xj, which will cost his spaceship xi×yj?xj×yi units of energy. Note that this cost can be negative, it means the flight will supply his spaceship.
Please write a program to help Little Q find the best route with minimum total cost.
In each test case, there is an integer n(2≤n≤200000) in the first line, denoting the number of planets.
For the next n lines, each line contains 2 integers xi,yi(0≤xi,yi≤109), denoting the coordinate of the i-th planet. Note that different planets may have the same coordinate because they are too close to each other. It is guaranteed that y1=yn=0,0=x1<x2,x3,...,xn?1<xn.
A sequence of integers a is lexicographically smaller than a sequence of b if there exists such index j that ai=bi for all i<j, but aj<bj.
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2018 Multi-University Training Contest 3 1007 / hdu6325 Problem G. Interstellar Travel 鍑稿寘