fzu 2039 Pets （简单二分图 + （最大流 || 二分图））

Posted 2020-09-29 lxjshuju

Are you interested in pets? There is a very famous pets shop in the center of the ACM city. There are totally m pets in the shop, numbered from 1 to m. One day, there are n customers in the shop, which are numbered from 1 to n. In order to sell pets to as more customers as possible, each customer is just allowed to buy at most one pet. Now, your task is to help the manager to sell as more pets as possible.

Every customer would not buy the pets he/she is not interested in it, and every customer would like to buy one pet that he/she is interested in if possible.

Input

There is a single integer T in the first line of the test data indicating that there are T(T≤100) test cases. In the first line of each test case, there are three numbers n, m(0≤n,m≤100) and e(0≤e≤n*m). Here, n and m represent the number of customers and the number of pets respectively.

In the following e lines of each test case, there are two integers x(1≤x≤n), y(1≤y≤m) indicating that customer x is not interested in pet y, such that x would not buy y.

Output

For each test case, print a line containing the test case number (beginning with 1) and the maximum number of pets that can be sold out.

Sample Input
1 2 2 2 1 2 2 1

Sample Output
Case 1: 2

题目大意：有n个顾客。有m仅仅宠物，而且顾客有e个要求。要求内容为。第i号顾客不想买第j号宠物。问最多能卖多少仅仅宠物。

解题思路：能够用最大流。能够用匈牙利hungary算法来求二分图。

最大流的时候。要注意拆点。建立一个超级源点连接全部的顾客，容量为INF。建立一个超级汇点使全部宠物连向他，容量为INF。

顾客和宠物各自拆成两个点，容量为1，这样能够保证，每一个顾客仅仅能买一仅仅宠物，每仅仅宠物仅仅能被一个顾客购买。然后依据e个要求，建立顾客和宠物之间的边，容量为1，之后求最大流。这种方法更复杂更耗时。所以这题最好用匈牙利算法。

最大流

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 1005;
const int OF1 = 100;
const int OF2 = 200;
const int FIN = 505;
const int INF = 0x3f3f3f3f;
int n, m, e, f[N][N], s, t;
struct Edge{
    int from, to, cap, flow; 
};

vector<Edge> edges;
vector<int> G[N];

void init() {
    s = 0, t = FIN;
    for (int i = 0; i < N; i++) G[i].clear();
    edges.clear();
    memset(f, 0, sizeof(f));
}

void addEdge(int from, int to, int cap, int flow) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    int temp = edges.size();
    G[from].push_back(temp - 2);
    G[to].push_back(temp - 1);
} 
void input() {
    int a, b;
    for (int i = 0; i < e; i++) {
        scanf("%d %d", &a, &b);
        f[a][b] = 1;
    }
    for (int i = 1; i <= n; i++) {
        addEdge(0, i, INF, 0);  
        addEdge(i, i + OF1, 1, 0);
    }
    for (int i = 1; i <= m; i++) {
        addEdge(i + OF2, i + OF2 + OF1, 1, 0);  
        addEdge(i + OF2 + OF1, FIN, INF, 0);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (!f[i][j]) {
                addEdge(i + OF1, j + OF2, 1, 0);        
            }
        }   
    }
}
int vis[N], d[N];
int BFS() {
    memset(vis, 0, sizeof(vis));
//  for (int i = 0; i < FIN; i++) d[N] = INF;
    queue<int> Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty()) {
        int u = Q.front(); Q.pop(); 
        for (int i = 0; i < G[u].size(); i++) {
            Edge &e = edges[G[u][i]];   
            if (!vis[e.to] && e.cap > e.flow) {
                vis[e.to] = 1;  
                d[e.to] = d[u] + 1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

int cur[N];
int DFS(int u, int a) {
    if (u == t || a == 0) return a;
    int flow = 0, f; 
    for (int &i = cur[u]; i < G[u].size(); i++) {
        Edge &e = edges[G[u][i]];
        if (d[u] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
            e.flow += f;    
            edges[G[u][i]^1].flow -= f;
            flow += f;
            a -= f;
            if (a == 0) break;
        }
    }
    return flow;
}
int MF() {
    int ans = 0;
    while (BFS()) {
        memset(cur, 0, sizeof(cur));    
        ans += DFS(s, INF);
    }
    return ans;
}
int main() {
    int T, Case = 1;
    scanf("%d", &T);
    while (T--) {
        printf("Case %d: ", Case++);
        scanf("%d %d %d", &n, &m, &e);  
        init();
        input();
        int ans = MF();
        printf("%d\n", ans);
    }
    return 0;
}

匈牙利算法

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef __int64 ll;
const int N = 505;
int n, m, e, ans;
int G[N][N], vis[N], R[N];
void input() {
    memset(G, 1, sizeof(G));
    memset(R, 0, sizeof(R));
    int a, b;
    for (int i = 0; i < e; i++) {
        scanf("%d %d", &a, &b); 
        G[a][b] = 0;
    }   
}
int find(int x) {
    for (int i = 1; i <= m; i++) {
        if (G[x][i] && !vis[i]) {
            vis[i] = 1;
            if (R[i] == 0 || find(R[i])) {
                R[i] = x;   
                return 1;
            }
        }
    }
    return 0;
}
void hungary() {
    for (int i = 1; i <= n; i++) {
        memset(vis, 0, sizeof(vis));
        if (find(i)) ans++;
    }
}
int main() {
    int T, Case = 1;
    scanf("%d", &T);    
    while (T--) {
        printf("Case %d: ", Case++);
        ans = 0;
        scanf("%d %d %d", &n, &m, &e);  
        input();
        hungary();
        printf("%d\n", ans);
    }
    return 0;
}