题目连接：http://poj.org/problem?

id=1087

题意：
n种插座 ，m个电器，f组（x，y）表示插座x能够替换插座y，问你最多能给几个电器充电。

解法：起点向插座建边，容量１，电器向汇点建边。容量１，插座向电器建边。容量１，能够替换的插座间建边。容量无穷大。然后套板子。

。。求最大流。

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>  using namespace std;const int MAXN = 1010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{int to, next, cap, flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];void init()
{tol = 0;memset(head, -1, sizeof(head));
}
//加边。单向图三个參数，双向图四个參数
void addedge(int u, int v, int w, int rw = 0)
{edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u];edge[tol].flow = 0; head[u] = tol++;edge[tol].to = u; edge[tol].cap = rw; edge[tol].next = head[v];edge[tol].flow = 0; head[v] = tol++;
}//输入參数：起点、终点、点的总数
//点的编号没有影响，仅仅要输入点的总数int sap(int start, int end, int N)
{memset(gap, 0, sizeof(gap));memset(dep, 0, sizeof(dep));memcpy(cur, head, sizeof(head));int u = start;pre[u] = -1;gap[0] = N;int ans = 0;while (dep[start] < N){if (u == end){int Min = INF;for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])if (Min > edge[i].cap - edge[i].flow)Min = edge[i].cap - edge[i].flow;for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to]){edge[i].flow += Min;edge[i ^ 1].flow -= Min;}u = start;ans += Min;continue;}bool flag = false;int v;for (int i = cur[u]; i != -1; i = edge[i].next){v = edge[i].to;if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]){flag = true;cur[u] = pre[v] = i;break;}}if (flag){u = v;continue;}int Min = N;for (int i = head[u]; i != -1; i = edge[i].next)if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){Min = dep[edge[i].to];cur[u] = i;}gap[dep[u]]--;if (!gap[dep[u]])return ans;dep[u] = Min + 1;gap[dep[u]]++;if (u != start) u = edge[pre[u] ^ 1].to;}return ans;
}int m, n, f;
map<string, int> Hash;
string x, y;int main()
{while (cin >> n){init();Hash.clear();int num1 = 2;int from = 0;int end = 1;for (int i = 1; i <= n; i++){cin >> x;Hash[x] = num1;addedge(0, num1, 1);num1++;}cin >> m;for (int i = 1; i <= m; i++){cin >> x >> y;if (Hash[x] == 0) Hash[x] = num1++;if (Hash[y] == 0) Hash[y] = num1++;addedge(Hash[x], end, 1);addedge(Hash[y], Hash[x], 1);}cin >> f;for (int i = 1; i <= f; i++){cin >> x >> y;if (Hash[x] == 0) Hash[x] = num1++;if (Hash[y] == 0) Hash[y] = num1++;addedge(Hash[y], Hash[x], 10000000);}int ans = sap(from, end, num1);printf("%d\n", m - ans);}return 0;
}