POJ 1087 A Plug for UNIX (网络流, Dinic 建图)

A Plug for UNIX

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17529 Accepted: 6055

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 bureaucratic as possible. 
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

The input will consist of one case. The first line contains a single positive integer n (1 <= n <= 100) indicating the number of receptacles in the room. The next n lines list the receptacle types found in the room. Each receptacle type consists of a string 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

A line containing a single non-negative integer indicating the smallest number of devices that cannot be plugged in.

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

这一道题, 建图  建了两天,    mmp 要疯了

题意:  n个插座, m个电器  k个转换器,   问有最少有几个 电器没有连,   求 网络流最大  然后 m-maxflow

主要问题在建图上,  因为有转换器  而且转换器无限制;

注意:1,  转换器转换的 可能存在插座没有的,

        2, 存在电器重复 插头 , 因此要查重建图,

       3 , 手动建立源点和汇点

类似一个这样的图:

套用Dinic  模板 

#include <iostream>#include <stdio.h>#include <algorithm>#include <math.h>#include <cmath>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt","w",stdout)#define S1(n)    scanf("%d",&n)#define SL1(n)   scanf("%I64d",&n)#define S2(n,m)  scanf("%d%d",&n,&m)#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)#define Pr(n)     printf("%d\n",n)using namespace std;typedef long long ll;const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int MOD=1e9+7;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};struct node{    int v,w,next; //u  v 从 u-v 权值为w}edge[maxn];int head[maxn],num[maxn],start,END,cnt,sum;int n,m,k;void add(int u,int v,int w){    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].w=0;    edge[cnt].next=head[v];    head[v]=cnt++;}int bfs(){    queue<int>Q;    mem(num,0);    num[start]=1;    Q.push(start);    while(!Q.empty())    {        int t=Q.front();        Q.pop();        if(t==END)            return 1;        for(int i=head[t];i!=-1;i=edge[i].next)// 链式前向星访问找增广路        {            int t1= edge[i].v;//下一个节点            int t2= edge[i].w;// 当前点 权值            if(t2&&num[t1]==0)// 当前点存在 并且下一个点没有访问            {                num[t1]=num[t]+1;// 点=1                if(t1==END)//  结束                    return 1;                Q.push(t1);            }        }    }    return 0;}int dfs(int u,int maxflow){    if(u==END)        return maxflow;    int res=0;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int t1=edge[i].v;// 下一个节点        int t2=edge[i].w;// 当前节点        if(t2&&num[t1]==num[u]+1)        {            int temp=dfs(t1,min(maxflow-res,t2));// 选择流 小的一部分            edge[i].w-=temp;// 正向减少            edge[i^1].w+=temp;//反向增加            res+=temp;            if(maxflow==res)                return res;        }    }    if(!res)        num[u]=-1;    return res;}void Dinic(){    int ans=0;    while(bfs())    {        ans+=dfs(start,INF);    }    cout<<m-ans<<endl;}map<string,int>maps;char sp[200],plug[200];int main(){    while(~S1(n))    {        mem(head,-1);        cnt=0;        maps.clear();        start=0,END=1000;        int num=1;        for(int i=1;i<=n;i++)        {            scanf("%s",sp);            maps[sp]=num++;            add(start,maps[sp],1);// 插座到 源点        }        S1(m);        for(int i=1;i<=m;i++)        {            scanf("%s %s",sp,plug);            maps[sp]=num++;            if(!maps[plug])// 没有                maps[plug]=num++;            add(maps[plug],maps[sp],1);            add(maps[sp],END,1);        }        S1(k);        for(int i=1;i<=k;i++)        {            scanf("%s %s",sp,plug);            if(!maps[sp])                maps[sp]=num++;            if(!maps[plug])                maps[plug]=num++;            add(maps[plug],maps[sp],INF);        }        Dinic();    }    return 0;}

123