POJ. 1005 I Think I Need a Houseboat(水 ) 代码总览 #include <cstdio> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <vector> #define nmax using namespace std; const double pi = acos(-1); int m…

I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 85149   Accepted: 36857 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land,…

1.链接地址: http://bailian.openjudge.cn/practice/1005/ http://poj.org/problem?id=1005 2.题目: I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 82376   Accepted: 35628 Description Fred Mapper is considering purchasing…

    I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 81874   Accepted: 35368 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the la…

一. 题目 I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 97512   Accepted: 42430 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the…

#include <iostream> using namespace std; const double pi = 3.1415926535; int main() { ;; double x,y; cin >> t; while(t--) { ++time; cin >> x >> y; double area = pi*(x*x + y*y); ; cout<<"Property "<<time<<…

一个公司的职员是分级制度的,所有员工刚好是一个树形结构,现在公司要举办一个聚会,邀请部分职员来参加. 要求: 1.为了聚会有趣,若邀请了一个职员,则该职员的直接上级(即父节点)和直接下级(即儿子节点)都不能被邀请 2.每一个员工都有一个兴奋值,在满足1的条件下,要使得邀请来的员工的兴奋值最高 输出最高的兴奋值. 简单的树形DP dp[i][1]:表示以i为根的子树,邀请节点i的最大兴奋值 dp[i][0]:表示以i为根的子树,不邀请节点i的最大兴奋值 先根据入度找出整棵树的根节点, 然后一次DF…

Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21010   Accepted: 10614 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertica…

#include <iostream> using namespace std; /*248K 32MS*/ int main() { int s,d; while(cin>>s>>d) { int count=0; for(int i=5;i>0;i--) { if(s*i<(5-i)*d) { count=i; break; } } int total=s*count-d*(5-count); total*=2; if(count>=2) tota…

id=1269" rel="nofollow">Intersecting Lines 大意:给你两条直线的坐标,推断两条直线是否共线.平行.相交.若相交.求出交点. 思路:线段相交推断.求交点的水题.没什么好说的. struct Point{ double x, y; } ; struct Line{ Point a, b; } A, B; double xmult(Point p1, Point p2, Point p) { return (p1.x-p.x)*(p2…