题意:题意:给出n个人的在x轴的位置和最大速度,求n个人相遇的最短时间。
析:二分时间,然后求并集,注意精度,不然会超时。
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-7; const int maxn = 60000 + 5; 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; } struct Node{ double x, v; }; Node a[maxn]; bool judge(double m){ double l = a[0].x-a[0].v*m, r = a[0].x+a[0].v*m; for(int i = 1; i < n; ++i){ l = max(l, a[i].x-a[i].v*m); r = min(r, a[i].x+a[i].v*m); if(r < l) return false; } return r >= l; } int main(){ scanf("%d", &n); for(int i = 0; i < n; ++i){ scanf("%lf", &a[i].x); } for(int i = 0; i < n; ++i){ scanf("%lf", &a[i].v); } double l = 0.0, r = 1e9; while(r - l > eps){ double m = (l + r) / 2.0; if(judge(m)) r = m; else l = m; } printf("%.6f\n", l); return 0; }