USACO: The Perfect Stall

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农场主约翰的新牛棚刚建成不久，但因为工程问题，每个产奶位大小不一。经过一周的数据收集，现在需要根据每头奶牛愿意在哪些产奶位产奶的数据，计算出最大的有效产奶位分配方案。

The Perfect Stall
Hal Burch

Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall.

Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.

PROGRAM NAME: stall4

INPUT FORMAT

Line 1:	One line with two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn.
Line 2..N+1:	N lines, each corresponding to a single cow. The first integer (S_i) on the line is the number of stalls that the cow is willing to produce milk in (0 <= S_i <= M). The subsequent S_i integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.

SAMPLE INPUT (file stall4.in)

OUTPUT FORMAT

A single line with a single integer, the maximum number of milk-producing stall assignments that can be made.

SAMPLE OUTPUT (file stall4.out)

解题思路：

最大二分匹配，几乎是原封不动地用了上一道题的代码

/* ID: geterns1 PROG: stall4 LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; //最大流Ford-Fulkerson算法，Edmonds-Karp优化 const int infi = 1073741824; const int size = 400 + 10; //顶点数 int realSize; int s, t; //源点和汇点 int edge[size][size]; //保存边的容量 int flow[size][size]; //保存两点流量 //初始化图和流 void initail(){ memset(edge, 0, sizeof(edge)); memset(flow, 0, sizeof(flow)); } // int root[size]; int list[2 * size], head, cnt; int ford_fulkerson(){ /* ... */ } int main(){ freopen("stall4.in", "r", stdin); freopen("stall4.out", "w", stdout); int n, m, c, d; scanf("%d %d", &n, &m); realSize = m + n + 2; initail(); for (int i = 1; i <= n; i ++){ scanf("%d", &c); while (c --){ scanf("%d", &d); edge[i][d + n] = 1; } } for (int i = 1; i <= n; i ++) edge[n + m + 1][i] = 1; for (int i = n + 1; i <= n + m; i ++) edge[i][n + m + 2] = 1; s = n + m + 1; t = n + m + 2; printf("%d/n", ford_fulkerson()); fclose(stdin); fclose(stdout); return 0; }