poj 2714 Random Walk

　　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！

View Code

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 
 7 using namespace std;
 8 const int maxn = 101;
 9 
10 struct vec{
11     double x;
12     double y;
13     bool operator < (const vec &a) const{
14         double t = atan2(y, x);
15         double at = atan2(a.y, a.x);
16 
17         return t < at;
18     }
19 }v[maxn << 1];
20 double maxx, maxy;
21 double curx, cury;
22 
23 int main(){
24     int n;
25     double x, y;
26 
27     while (~scanf("%d", &n) && n){
28         for (int i = 0; i < n; i++){
29             scanf("%lf%lf", &x, &y);
30             v[i << 1].x = x;
31             v[i << 1].y = y;
32             v[i << 1 | 1].x = -x;
33             v[i << 1 | 1].y = -y;
34         }
35         sort(v, v + (n << 1));
36         curx = cury = 0;
37         for (int i = 0; i < n; i++){
38             curx += v[i].x;
39             cury += v[i].y;
40         }
41         maxx = curx;
42         maxy = cury;
43         for (int i = 0, end = n << 1; i < end; i++){
44             int ci = (i + n) % end;
45 
46             curx += v[ci].x - v[i].x;
47             cury += v[ci].y - v[i].y;
48             if (curx * curx + cury * cury > maxx * maxx + maxy * maxy)
49                 maxx = curx, maxy = cury;
50         }
51 
52         printf("Maximum distance = %.3f meters.\n", sqrt((double)maxx * maxx + maxy * maxy));
53     }
54 
55     return 0;
56 }

——written by Lyon