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…

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…

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…

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…

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…

HDU 3549 Flow Problem(最大流) Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) [Description] [题目描述] Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for t…

HDU 4974 A simple water problem pid=4974" target="_blank" style="">题目链接 签到题,非常easy贪心得到答案是(sum + 1) / 2和ai最大值的最大值 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N =…

hdu 6214 Smallest Minimum Cut[最大流] 题意:求最小割中最少的边数. 题解:对边权乘个比边大点的数比如300,再加1 ,最后,最大流对300取余就是边数啦.. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue…