hdoj 1526 A Plug for UNIX 【floyd求传递闭包 + 二分图匹配】

A Plug for UNIX

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 892    Accepted Submission(s): 200


Problem 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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.
 

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
 
   
1 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
 

题意:有N个插座,现在有M个人且每个人都有需要使用的插座。又给你K种转换器即可以把插座A变成插座B(可以无限使用)。问你最少有几个人用不到插座。

 

转换问题:把已有的N个插座当作N个男生(根据自反性建图),人所需要的M个插座当作M个女生。

 

思路:用map建立字符串和它出现编号的映射,根据字符串对应编号建图。对于K个转换器给出的信息,可以虚拟成K条单向边。由于转换器可以无限使用,因此需要先求出传

递闭包(节点数目为出现的所有插座的数目)。最后再用匈牙利算法求出M个女生的最大匹配数(每个女生的匹配范围为N个男生),用M减去它即可。

 

注意:M个人和转换器里面可能出现新的插座。。。坑死。还有格式每两个测试数据有一个空格,最后一个测试数据没有空格。

 

#include 
#include 
#include 
#include 
#include 
#define MAXN 1000+10
using namespace std;
int Map[MAXN][MAXN];
int pipei[MAXN];
bool used[MAXN];
int Rnum;//已有的插座数目
int Mnum;//使用插座的人数
map KK;
int RR[MAXN];//存储已有插座 
int MM[MAXN];//存储每个人需要插座的编号 
int num;//记录出现插座的总数目
void input()
{
    char a[110], b[110];
    memset(Map, 0, sizeof(Map));
    KK.clear();//清空
    num = 1;
    scanf("%d", &Rnum);
    for(int i = 1; i <= Rnum; i++)
    {
        scanf("%s", a);
        if(!KK[a]) KK[a] = num++;//建立映射
        RR[i] = KK[a];
        Map[RR[i]][RR[i]] = 1;//自己连接自己 
    } 
    scanf("%d", &Mnum);
    for(int i = 1; i <= Mnum; i++)
    {
        scanf("%s%s", a, b);
        if(!KK[b]) KK[b] = num++;
        MM[i] = KK[b];//第i个人对应插座编号
    }
    int k;
    scanf("%d", &k);
    while(k--)
    {
        scanf("%s%s", &a, &b);//可相互转换的插座 
        if(!KK[a]) KK[a] = num++;
        if(!KK[b]) KK[b] = num++;
        Map[KK[a]][KK[b]] = 1;//建立关系
    }
}
void Floyd()//求传递闭包
{
    for(int k = 1; k < num; k++)//以所有出现的插座数目为上限 求传递闭包 
    {
        for(int i = 1; i < num; i++)
        {
            for(int j = 1; j < num; j++)
            {
                if(Map[i][k] && Map[k][j])
                    Map[i][j] = 1;
            }
        }
    }
}
int find(int x)//以第x个人需要使用的 插座编号来匹配 
{
    for(int i = 1; i <= Rnum; i++)//依次匹配Rnum个已有插座 
    {
        if(Map[MM[x]][RR[i]] && !used[RR[i]])
        {
            used[RR[i]] = true;
            if(pipei[RR[i]] == -1 || find(pipei[RR[i]]))
            {
                pipei[RR[i]] = x;
                return 1;
            }
        }
    }
    return 0;
}
void solve()
{
    Floyd();//求传递闭包
    memset(pipei, -1, sizeof(pipei));
    int ans = 0;
    for(int i = 1; i <= Mnum; i++)//对Mnum个人求匹配 就是对他所对应的插座匹配
    {
        memset(used, false, sizeof(used));
        //printf("%d\n", MM[i]);
        ans += find(i);
    }
    printf("%d\n", Mnum - ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        input();
        solve();
        if(t)
        printf("\n"); 
    }
    return 0;
}


 

 

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