CodeForces 703C Chris and Road (简单几何)
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题意:有一个n边形的汽车向以速度v向x轴负方向移动,给出零时时其n个点的坐标。并且有一个人在(0,0)点,可以以最大速度u通过w宽的马路,到达(0,w)点。现在要求人不能碰到汽车,人可以自己调节速度。问人到达马路对面的最小时间是多少?
析:这个题是一个简单的分类讨论,很明显只有两种情况,第一种,直接到达w,不会被车撞到,答案就是w/u,
第二种是切着车过去,或者是等车过去再过去,只要枚举车的每个顶点,找到最后通过y轴的点就好,或者根本不会与车相切。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2e5 + 10; const int mod = 1000000007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int main(){ int w, u, v; scanf("%d %d %d %d", &n, &w, &v, &u); bool ok = true; double ans = 0.0; for(int i = 0; i < n; ++i){ int x, y; scanf("%d %d", &x, &y); double t1 = x * 1.0 / v; double t2 = y*1.0 / u; if(t1 < t2) ok = false; ans = max(ans,t1 + (w-y)*1.0 / u); } if(ok) printf("%.6f\n", w*1.0/u); else printf("%.6f\n", max(ans, w*1.0/u)); return 0; }
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