poj 2714 Random Walk

http://poj.org/problem?id=2714

　　因为每个向量的方向都不同，所以就可以将向量按圆周顺序排一下序，然后枚举每半个圆周的向量和，找出最大值就行了。

　　这里是用了贪心的思想，尽量将向量放到同一侧，这样就可以尽可能大的构造一个和向量！

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <cstring>

using namespace std;
const int maxn = 101;

struct vec{
    double x;
    double y;
    bool operator < (const vec &a) const{
        double t = atan2(y, x);
        double at = atan2(a.y, a.x);

        return t < at;
    }
}v[maxn << 1];
double maxx, maxy;
double curx, cury;

int main(){
    int n;
    double x, y;

    while (~scanf("%d", &n) && n){
        for (int i = 0; i < n; i++){
            scanf("%lf%lf", &x, &y);
            v[i << 1].x = x;
            v[i << 1].y = y;
            v[i << 1 | 1].x = -x;
            v[i << 1 | 1].y = -y;
        }
        sort(v, v + (n << 1));
        curx = cury = 0;
        for (int i = 0; i < n; i++){
            curx += v[i].x;
            cury += v[i].y;
        }
        maxx = curx;
        maxy = cury;
        for (int i = 0, end = n << 1; i < end; i++){
            int ci = (i + n) % end;

            curx += v[ci].x - v[i].x;
            cury += v[ci].y - v[i].y;
            if (curx * curx + cury * cury > maxx * maxx + maxy * maxy)
                maxx = curx, maxy = cury;
        }

        printf("Maximum distance = %.3f meters.\n", sqrt((double)maxx * maxx + maxy * maxy));
    }

    return 0;
}

——written by Lyon