HDU 4036 物理坑爹题-白红宇

HDU 4036 物理坑爹题

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发布时间：2019-05-10

本文共 2504 字，大约阅读时间需要 8 分钟。

/*******************************************************************************   坑爹的地方比较多，首先所有质量都是没用的，因为无摩擦力；其次坐标最好都用double表示；还有那个土豆是的坐标是相对于第一个peak的坐标，第四貌似有土豆可以不在peak的范围内。。。以后找min/max值还是用第一个元素做标记比较好。。。*******************************************************************************/#include 
   
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                           using namespace std;#define LOWBIT(x) ( (x) & ( (x) ^ ( (x) - 1 ) ) )#define CLR(x, k) memset((x), (k), sizeof(x))#define CPY(t, s) memcpy((t), (s), sizeof(s))#define SC(t, s) static_cast
                           
                            (s)#define LEN(s) static_cast
                            
                             ( strlen((s)) )#define SZ(s) static_cast
                             
                              ( (s).size() )typedef double LF;typedef __int64 LL; //VCtypedef unsigned __int64 ULL;typedef pair
                              
                                PII;typedef pair
                               
                                 PLL;typedef pair
                                
                                  PDD;typedef vector
                                 
                                   VI;typedef vector
                                  
                                    VC;typedef vector
                                   
                                     VF;typedef vector
                                    
                                      VS;template 
                                     
                                      T sqa(const T & x){ return x * x;}template 
                                      
                                       T gcd(T a, T b){ if (!a || !b) { return max(a, b); } T t; while (t = a % b) { a = b; b = t; } return b;};const int INF_INT = 0x3f3f3f3f;const LL INF_LL = 0x7fffffffffffffffLL; //15fconst double oo = 10e9;const double eps = 10e-7;const double PI = acos(-1.0);#define ONLINE_JUDGEconst int MAXN = 10004;int test, n, m, w;struct Peak{ LF x, h;}peak[MAXN];struct Point{ LF x, v;}pnt[MAXN];void ace(){ int cas = 1; LF ans = 0; for (cin >> test; test--; ++cas) { cin >> n >> m >> w; LF maxh = peak[0].h; for (int i = 0; i < n; ++i) { cin >> peak[i].x >> peak[i].h; maxh = max(maxh, peak[i].h); } ans = sqrt(2.0 * 20 * (maxh - peak[0].h)); for (int i = 0; i < m; ++i) { int tw; cin >> pnt[i].x >> pnt[i].v >> tw; pnt[i].x += peak[0].x; for (int k = 0; k < n - 1; ++k) { if (peak[k].x < pnt[i].x + eps && pnt[i].x < peak[k + 1].x) { LF dh = peak[k + 1].h - peak[k].h; LF dx = peak[k + 1].x - peak[k].x; LF ph = peak[k].h + (pnt[i].x - peak[k].x) / dx * dh; LF buf = pnt[i].v * pnt[i].v + 2.0 * 20 * (ph - peak[0].h); if (buf + eps > 0) { ans = max(ans, sqrt(buf)); } break ; } } } printf("Case %d: %.2lf\n", cas, ans); } return ;}int main(){#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif ace(); return 0;}/*******************************************************************************Test Data...16 2 1000 02 53 24 15 38 -22 15 1005 11 100*******************************************************************************/