POJ 1087 A Plug for UNIX 最大流

看完题以后就觉得是个最大流了。然后脑子不太清醒，想了一会儿才发现建图好简单啊，不过中间理解错题意了，题目中说有K种适配器，也就是说每种适配器的数量是无限个的，而我理解成了K个适配器了。。。 然后就WA了几次。。。

建图：  加一个超级源点， 超级汇点。 

源点与每个电器相连，容量为1，每个电器与其相应的插座相连，容量为1，插座之间能转化的就连一条无限容量的边，最后每个插座跟汇点相连，容量为这种类型的插座的个数。

需要注意的是，最后图中的节点数可能会达到400之多。题目的数据可能会给出一些看起来毫无用处的适配器，比如把一个数量为0的插座类型转化为另一种数量为0的插座类型。 这个也是要计算在内的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdlib>
#include <map>
#include <algorithm>
#define MAXN 505
#define MAXM 500005
#define INF 1111111111
using namespace std;
struct node
{
    int ver;    // vertex
    int cap;    // capacity
    int flow;   // current flow in this arc
    int next, rev;
}edge[MAXM];
int dist[MAXN], numbs[MAXN], src, des, n;
int head[MAXN], e;
void add(int x, int y, int c)
{       //e记录边的总数
    //printf("%d %d %d\n", x, y, c);
    edge[e].ver = y;
    edge[e].cap = c;
    edge[e].flow = 0;
    edge[e].rev = e + 1;        //反向边在edge中的下标位置
    edge[e].next = head[x];   //记录以x为起点的上一条边在edge中的下标位置
    head[x] = e++;           //以x为起点的边的位置
    //反向边
    edge[e].ver = x;
    edge[e].cap = 0;  //反向边的初始网络流为0
    edge[e].flow = 0;
    edge[e].rev = e - 1;
    edge[e].next = head[y];
    head[y] = e++;
}
void rev_BFS()
{
    int Q[MAXN], qhead = 0, qtail = 0;
    for(int i = 1; i <= n; ++i)
    {
        dist[i] = MAXN;
        numbs[i] = 0;
    }
    Q[qtail++] = des;
    dist[des] = 0;
    numbs[0] = 1;
    while(qhead != qtail)
    {
        int v = Q[qhead++];
        for(int i = head[v]; i != -1; i = edge[i].next)
        {
            if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;
            dist[edge[i].ver] = dist[v] + 1;
            ++numbs[dist[edge[i].ver]];
            Q[qtail++] = edge[i].ver;
        }
    }
}
void init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}
int maxflow()
{
    int u, totalflow = 0;
    int Curhead[MAXN], revpath[MAXN];
    for(int i = 1; i <= n; ++i)Curhead[i] = head[i];
    u = src;
    while(dist[src] < n)
    {
        if(u == des)     // find an augmenting path
        {
            int augflow = INF;
            for(int i = src; i != des; i = edge[Curhead[i]].ver)
                augflow = min(augflow, edge[Curhead[i]].cap);
            for(int i = src; i != des; i = edge[Curhead[i]].ver)
            {
                edge[Curhead[i]].cap -= augflow;
                edge[edge[Curhead[i]].rev].cap += augflow;
                edge[Curhead[i]].flow += augflow;
                edge[edge[Curhead[i]].rev].flow -= augflow;
            }
            totalflow += augflow;
            u = src;
        }
        int i;
        for(i = Curhead[u]; i != -1; i = edge[i].next)
            if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;
        if(i != -1)     // find an admissible arc, then Advance
        {
            Curhead[u] = i;
            revpath[edge[i].ver] = edge[i].rev;
            u = edge[i].ver;
        }
        else        // no admissible arc, then relabel this vertex
        {
            if(0 == (--numbs[dist[u]]))break;    // GAP cut, Important!
            Curhead[u] = head[u];
            int mindist = n;
            for(int j = head[u]; j != -1; j = edge[j].next)
                if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);
            dist[u] = mindist + 1;
            ++numbs[dist[u]];
            if(u != src)
                u = edge[revpath[u]].ver;    // Backtrack
        }
    }
    return totalflow;
}
int num[505];
int main()
{
    int nt, m, k;
    string str1, str2;
    map<string ,int>plug;
    init();
    scanf("%d", &nt);
    int cnt = 0;
    for(int i = 0; i < nt; i++)
    {
        cin >> str1;
        if(plug[str1] == 0) plug[str1] = ++cnt;
        num[plug[str1]]++;
    }
    scanf("%d", &m);
    int t = cnt;
    src = 1;
    for(int i = 0; i < m; i++)
    {
        cin >> str1 >> str2;
        if(plug[str2] == 0) plug[str2] = ++cnt;
        add(2 + i, plug[str2] + m + 1, 1);
        add(src, 2 + i, 1);
    }
    scanf("%d", &k);
    for(int i = 0; i < k; i++)
    {
        cin >> str1 >> str2;
        if(plug[str1] == 0) plug[str1] = ++cnt;
        if(plug[str2] == 0) plug[str2] = ++cnt;
        add(plug[str1] + m + 1, plug[str2] + m + 1, INF);
    }
    n = 2 + m + cnt;
    des = n;
    for(int i = 1; i <= t; i++) add(m + 1 + i, des, num[i]);
    //printf("ss %d %d\n", src, des);
    rev_BFS();
    int ans = m - maxflow();
    printf("%d\n", ans);
    return 0;
}