fzu 2039 Pets （簡單二分圖 + （最大流 || 二分圖）），fzu2039

fzu 2039 Pets （簡單二分圖 + （最大流 || 二分圖）），fzu2039

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

題目大意：有n個顧客，有m只寵物，並且顧客有e個要求，要求內容為，第i號顧客不想買第j號寵物。問最多能賣多少只寵物。解題思路：可以用最大流，可以用匈牙利hungary演算法來求二分圖。最大流的時候，要注意拆點。建立一個超級源點串連所有的顧客，容量為INF，建立一個超級匯點使所有寵物連向他，容量為INF。顧客和寵物各自拆成兩個點，容量為1，這樣可以保證，每個顧客只能買一隻寵物，每隻寵物只能被一個顧客購買。然後根據e個要求，建立顧客和寵物之間的邊，容量為1，之後求最大流。這個方法更複雜更耗時，所以這題最好用匈牙利演算法。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;}

匈牙利演算法

#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;}

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