C - A Plug for UNIX    You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumb…

A Plug for UNIX   Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet…

A Plug for UNIX 题目链接:https://vjudge.net/problem/POJ-1087 Description: You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow o…

题目链接:https://vjudge.net/problem/POJ-1087 A Plug for UNIX Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17861   Accepted: 6172 Description You are in charge of setting up the press room for the inaugural meeting of the United Nations In…

解题报告 题意: n个插头m个设备k种转换器.求有多少设备无法插入. 思路: 定义源点和汇点,源点和设备相连,容量为1. 汇点和插头相连,容量也为1. 插头和设备相连,容量也为1. 可转换插头相连,容量也为inf(由于插头有无限个) #include <map> #include <queue> #include <cstdio> #include <vector> #include <cstring> #include <iostream…

链接 : http://acm.hust.edu.cn/vjudge/problem/viewProblem.action? id=26746 题目意思有点儿难描写叙述 用一个别人描写叙述好的. 我的建图方法:一个源点一个汇点,和全部种类的插座.输入的n个插座直接与源点相连,容量为1,m个物品输入里 记录每一个插座相应的物品个数.物品数然后大于0的插座直接连到汇点.意味着终于的物品仅仅能由这些插座流出.中间的插座转换容量都是INF  a b表示  不管多少b都能够选择转化到a. /*------…

描述 You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and bureaucra…

Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and…

最大流. 流可以对应一种分配方式. 显然最大流就可以表示最多匹配数 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; + ; + ; ; const int inf = 0x3f3f3f3f; char s[maxn][maxl],t[maxl],t2[maxl]; int g[maxn],v[maxm],nex[maxm],f[maxm],eid; int id[ma…

题意:给定 n 种插座,m种设备,和k个转换器,问你最少有几台设备不能匹配. 析:一个很裸的网络流,直接上模板就行,建立一个源点s和汇点t,源点和每个设备连一条边,每个插座和汇点连一条边,然后再连转换器, 最后跑一次最大流即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib…