Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 629    Accepted Submission(s): 288 Problem Description Pahom on Water is an interactive computer game inspired by a short story of…

Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 770    Accepted Submission(s): 353 Problem Description Pahom on Water is an interactive computer game inspired by a short story of…

Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 772    Accepted Submission(s): 355 Problem Description Pahom on Water is an interactive computer game inspired by a short story of…

https://vjudge.net/problem/HDU-4183 题意: 这道题目的英文实在是很难理解啊. 给出n个圆,每个圆有频率,x.y轴和半径r4个属性,每次将频率为400的圆作为起点,频率为789点作为终点.从源点到汇点时必须从频率小的到频率大的,而从汇点到源点时必须从频率大的到频率小的.前提时这两个圆必须严格相交.每个点只能走一次.判断是否能从起点出发到达终点,并再次返回起点. 思路: 其实就是判断最大流是否大于等于2.因为每个点只能走一次,用拆点法. #include<iost…

题意: 有n个圆,每个圆的中心和半径和一个频率都给定,只有一个频率最高的789为紫色,只有一个最低的400为红色,规则如下: 1.当两个圆严格相交时,且人是从红色到紫色的方向运动时可以由低频率向高频率移动 2.当两个圆严格相交时,且人是从紫色到红色的方向运动时可以由高频率向低频率运动 3.除了红色的圆以外,离开某个圆之后就会消失(即只能走一次) 思路: 如果一开始红色和紫色就相交,则存在合理方案.否则 本题要求是先从红点出发,经过紫点之后再返回红点,如果以红点作为源点,网络流算法不能先到达一个T…

就是一个网络流.red结点容量为2,查看最大流量是否大于等于2.对于条件2,把边反向加入建图.条件1,边正向加入建图. /* 4183 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque>…

欢迎参加——每周六晚的BestCoder(有米!) Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 678    Accepted Submission(s): 312 Problem Description Pahom on Water is an interactive computer game ins…

Pahom on Water Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 885    Accepted Submission(s): 409 Problem Description Pahom on Water is an interactive computer game inspired by a short story of…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=4183 这题题目意思非常难看懂..我看了好长时间也没看懂..终于是从网上找的翻译. .我就在这翻译一下吧. 意思大约是:有多个点,每一个点给出坐标与半径,增加两个点相交,就能够从这两个点走.题目要求先从起点到终点,再从终点回到起点.从起点到终点的过程中.仅仅能从频率小的走到频率大的点(前提是两点相交).从终点到起点的过程中.仅仅能从频率大的走到频率小的.在走的过程中,除了起点与终点,别的仅仅要走过就会…

Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges. Wate…