USACO Section 4.2: The Perfect Stall

这题关键就在将题转换成最大流模板题。首先有一个原始点，N个cow个点， M个barn点和一个终点，原始点到cow点和barn点到终点的流都为1，而cow对应的barn就是cow点到对应barn点的流，为1.这样题目就转换成了原始点到终点的最大流问题

 1 /*

 2 ID: yingzho1

 3 LANG: C++

 4 TASK: stall4

 5 */

 6 #include <iostream>

 7 #include <fstream>

 8 #include <string>

 9 #include <map>

10 #include <vector>

11 #include <set>

12 #include <algorithm>

13 #include <stdio.h>

14 #include <queue>

15 #include <cstring>

16 #include <cmath>

17 #include <list>

18 #include <cstdio>

19 #include <cstdlib>

20 #include <limits>

21 #include <stack>

22 

23 using namespace std;

24 

25 ifstream fin("stall4.in");

26 ofstream fout("stall4.out");

27 

28 const int MAX = 210;

29 const int INF = 2000000000;

30 

31 int N, M;

32 int g[2*MAX][2*MAX], f[2*MAX][2*MAX], pre[2*MAX], inc[2*MAX];

33 

34 bool bfs(int s, int d) {

35     queue<int> que;

36     for (int i = 1; i <= N+M+2; i++) pre[i] = -1;

37     que.push(s);

38     inc[s] = INF;

39     while (!que.empty()) {

40         int u = que.front();

41         que.pop();

42         for (int i = 1; i <= N+M+2; i++) {

43             if (pre[i] == -1 && f[u][i] < g[u][i]) {

44                 inc[i] = min(inc[u], g[u][i]-f[u][i]);

45                 pre[i] = u;

46                 if (i == d) return true;

47                 que.push(i);

48             }

49         }

50     }

51     return false;

52 }

53 

54 int edmond_karp(int s, int d) {

55     int maxflow = 0;

56     while (bfs(s, d)) {

57         maxflow += inc[d];

58         for (int i = d; i != s; i = pre[i]) {

59             f[pre[i]][i] += inc[d];

60             f[i][pre[i]] -= inc[d];

61         }

62     }

63     return maxflow;

64 }

65 

66 int main()

67 {

68     fin >> N >> M;

69     int s, d;

70     for (int i = 1; i <= N; i++) {

71         g[1][1+i] = 1;

72         fin >> s;

73         for (int j = 0; j < s; j++) {

74             fin >> d;

75             g[1+i][1+N+d] = 1;

76         }

77     }

78     for (int i = N+2; i <= N+M+1; i++) g[i][N+M+2] = 1;

79     fout << edmond_karp(1, N+M+2) << endl;

80 

81     return 0;

82 }