UVa753/POJ1087_A Plug for UNIX(网络流最大流)(小白书图论专题)
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>
#include <algorithm>
#define inf 99999999
using namespace std;
int n,m,k,cnt,l[500],mmap[500][500],N[1000];
map<string,int >Map;
int bfs()
{
memset(l,-1,sizeof(l));
queue<int>Q;
Q.push(0);
l[0]=0;
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int i=0;i<=cnt;i++)
{
if(l[i]==-1&&mmap[u][i])
{
l[i]=l[u]+1;
Q.push(i);
}
}
}
if(l[cnt]>0)return 1;
return 0;
}
int dfs(int x,int f)
{
if(x==cnt)return f;
int a;
for(int i=0;i<=cnt;i++)
{
if(mmap[x][i]&&l[i]==l[x]+1&&(a=dfs(i,min(f,mmap[x][i]))))
{
mmap[x][i]-=a;
mmap[i][x]+=a;
return a;
}
}
return 0;
}
int main()
{
int i,j;
string a,b;
cin>>n;
cnt=1;
for(i=0;i<n;i++)
{
cin>>a;
N[cnt]=1;//cha
Map[a]=cnt++;
}
cin>>m;
for(i=0;i<m;i++)
{
cin>>a>>b;
N[cnt]=2;//she
Map[a]=cnt++;
if(Map[b]==0)
{
N[cnt]=3;//new1
Map[b]=cnt++;
}
int u=Map[a];
int v=Map[b];
mmap[u][v]=1;
}
cin>>k;
for(i=0;i<k;i++)
{
cin>>a>>b;
if(Map[a]==0)
{
N[cnt]=3;
Map[a]=cnt++;
}
if(Map[b]==0)
{
N[cnt]=3;
Map[b]=cnt++;
}
int u=Map[a];
int v=Map[b];
mmap[u][v]=inf;
}
for(i=1;i<cnt;i++)
{
if(N[i]==1)
{
mmap[i][cnt]=1;
}
if(N[i]==2)
{
mmap[0][i]=1;
}
}
int ans=0,t;
while(bfs())
while(t=dfs(0,inf))
ans+=t;
cout<<m-ans<<endl;
return 0;
}
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13550 | Accepted: 4518 |
Description
Since the room was designed to accommodate reporters and journalists from around the world, it is equipped with electrical receptacles to suit the different shapes of plugs and voltages used by appliances in all of the countries that existed when the room was
built. Unfortunately, the room was built many years ago when reporters used very few electric and electronic devices and is equipped with only one receptacle of each type. These days, like everyone else, reporters require many such devices to do their jobs:
laptops, cell phones, tape recorders, pagers, coffee pots, microwave ovens, blow dryers, curling
irons, tooth brushes, etc. Naturally, many of these devices can operate on batteries, but since the meeting is likely to be long and tedious, you want to be able to plug in as many as you can.
Before the meeting begins, you gather up all the devices that the reporters would like to use, and attempt to set them up. You notice that some of the devices use plugs for which there is no receptacle. You wonder if these devices are from countries that didn't
exist when the room was built. For some receptacles, there are several devices that use the corresponding plug. For other receptacles, there are no devices that use the corresponding plug.
In order to try to solve the problem you visit a nearby parts supply store. The store sells adapters that allow one type of plug to be used in a different type of outlet. Moreover, adapters are allowed to be plugged into other adapters. The store does not have
adapters for all possible combinations of plugs and receptacles, but there is essentially an unlimited supply of the ones they do have.
Input
of at most 24 alphanumeric characters. The next line contains a single positive integer m (1 <= m <= 100) indicating the number of devices you would like to plug in. Each of the next m lines lists the name of a device followed by the type of plug it uses (which
is identical to the type of receptacle it requires). A device name is a string of at most 24 alphanumeric
characters. No two devices will have exactly the same name. The plug type is separated from the device name by a space. The next line contains a single positive integer k (1 <= k <= 100) indicating the number of different varieties of adapters that are available.
Each of the next k lines describes a variety of adapter, giving the type of receptacle provided by the adapter, followed by a space, followed by the type of plug.
Output
Sample Input
4
A
B
C
D
5
laptop B
phone C
pager B
clock B
comb X
3
B X
X A
X D
Sample Output
1
UVa753/POJ1087_A Plug for UNIX(网络流最大流)(小白书图论专题)的更多相关文章
- UVa563_Crimewave(网络流/最大流)(小白书图论专题)
解题报告 思路: 要求抢劫银行的伙伴(想了N多名词来形容,强盗,贼匪,小偷,sad.都认为不合适)不在同一个路口相碰面,能够把点拆成两个点,一个入点.一个出点. 再设计源点s连向银行位置.再矩阵外围套 ...
- uva753 A Plug for UNIX 网络流最大流
C - A Plug for UNIX You are in charge of setting up the press room for the inaugural meeting of t ...
- poj 1087 C - A Plug for UNIX 网络流最大流
C - A Plug for UNIXTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contes ...
- UVA 753 - A Plug for UNIX(网络流)
A Plug for UNIX You are in charge of setting up the press room for the inaugural meeting of the U ...
- POJ1087 A Plug for UNIX 【最大流】
A Plug for UNIX Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13855 Accepted: 4635 ...
- POJ 1087 A Plug for UNIX (网络流,最大流)
题面 You are in charge of setting up the press room for the inaugural meeting of the United Nations In ...
- 【uva753/poj1087/hdu1526-A Plug for UNIX】最大流
题意:给定n个插座,m个插头,k个转换器(x,y),转换器可以让插头x转成插头y.问最少有多少个插头被剩下. 题解: 最大流或者二分图匹配.然而我不知道怎么打二分图匹配..打了最大流.这题字符串比较坑 ...
- poj 1087.A Plug for UNIX (最大流)
网络流,关键在建图 建图思路在代码里 /* 最大流SAP 邻接表 思路:基本源于FF方法,给每个顶点设定层次标号,和允许弧. 优化: 1.当前弧优化(重要). 1.每找到以条增广路回退到断点(常数优化 ...
- uva753 A Plug for UNIX
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<cstri ...
随机推荐
- poj 3259(bellman最短路径)
Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 30169 Accepted: 10914 Descr ...
- Jquery中使用setInterval和setTimeout 容易犯的低级错误
直接在ready中调用其他方法,会提示缺少对象的错误,解决方法如下: 方法1. 应用jQuery的扩展可以解决这个问题. $(document).ready(function(){ $.extend( ...
- MYSQL,innodb_buffer_pool_size内存分配
为MYSQL.innodb_buffer_pool_size=8G.MySQL一起动就会将占用掉8G内存(觉得TOP能够看到内存被使用了8G),可是近期才细致研究一下.原来不是这种(可能自己对Linu ...
- 15一个NoSql数据库
随着因特网web2.0该网站的兴起.非关系型数据库,现在已经成为一个非常受欢迎的新领域.非关系数据库产品的发展非常迅速.而在处理传统的关系数据库web2.0现场.特别是大规模,高并发SNS类型web2 ...
- RH033读书笔记(7)-Lab 8 Introduction to String Processing
Lab 8 Introduction to String Processing Sequence 1: Exercises in string processing 1. Other than the ...
- 杭电1171 Big Event in HDU(母函数+多重背包解法)
Big Event in HDU Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- FTP上传文件时 System.Net.WebException: 基础连接已经关闭: 接收时发生错误。
在c#中使用HttpWebRequest时,频繁请求一个网址时,过段时间就会出现“基础连接已经关闭: 接收时发生意外错误”的错误提示.将webRequest的属性设置成下面的,经测试可以解决.Syst ...
- 参加persist.sys物业写权限的方法
1.于AndroidManifest.xml manifest添加属性android:sharedUserId="android.uid.system" 2.假设AndroidMa ...
- 【SSH进阶之路】一步步重构容器实现Spring框架——彻底封装,实现简单灵活的Spring框架(十一)
文件夹 [SSH进阶之路]一步步重构容器实现Spring框架--从一个简单的容器開始(八) [SSH进阶之路]一步步重构容器实现Spring框架--解决容器对组件的"侵入 ...
- Swift 书面 ToDo App
下面的代码是使用的全部Xcode Version 6.0.1 (6A317)书面. 因为当使用团队开发stroyboard在并购的诸多不便的时间,所有或使用.xib该文件准备ToDo App. 想要实 ...