poj 1087 A Plug for UNIX 【最大流】

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可以替换插座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，可以替换的插座间建边，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，单向图三个参数，双向图四个参数
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++;
}

//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数

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;
}