UVa Problem 10310 Dog and Gopher （狗拿地鼠）

// Dog and Gopher （狗拿地鼠）
// PC/UVa IDs: 111301/10310, Popularity: A, Success rate: average Level: 1
// Verdict: Accepted
// Submission Date: 2011-11-01
// UVa Run Time: 0.044s
//
// 版权所有（C）2011，邱秋。metaphysis # yeah dot net
//
// [解题方法]
// 简单题，判断输入中那个洞与狗的距离大于与地鼠的距离两倍以上。#include <iostream>
#include <cmath>using namespace std;int main(int ac, char *av[])
{int n;double gopherX, gopherY, dogX, dogY, holeX, holeY;double emptyX, emptyY;while (cin >> n >> gopherX >> gopherY >> dogX >> dogY){bool successed = false;for (int i = 1; i <= n; i++){cin >> holeX >> holeY;// 判断距离。if ((pow(dogX - holeX, 2) + pow(dogY - holeY, 2)) >=4.0 * (pow(gopherX - holeX, 2) +pow(gopherY - holeY, 2))){successed = true;// 读取完剩余数据。for (int j = i + 1; j <= n; j++)cin >> emptyX >> emptyY;break;}}if (successed){cout.precision(3);cout.setf(ios::fixed | ios::showpoint);cout << "The gopher can escape through the hole at (";cout << holeX << "," << holeY << ")." << endl;}elsecout << "The gopher cannot escape." << endl;}return 0;
}

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UVa Problem 10310 Dog and Gopher （狗拿地鼠）

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