CodeForces 782B The Meeting Place Cannot Be Changed (二分)

时间：2022-08-30 08:10:04

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;

}

　　

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