2019独角兽企业重金招聘Python工程师标准>>> 
The Perfect Stall
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16888 | Accepted: 7721 |
Description
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.
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.
Input
The input includes several cases. For each case, the first line contains 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. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si 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.
Output
For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.
Sample Input
5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2Sample Output
4Source
USACO 40
我的解答:
二分图匹配.匈牙利算法.
/*=============================================================================
# FileName: 1274.cpp
# Desc: poj 1274
# Author: zhuting
# Email: cnjs.zhuting@gmail.com
# HomePage: my.oschina.net/locusxt
# Version: 0.0.1
# CreatTime: 2013-12-07 15:54:43
# LastChange: 2013-12-07 15:54:43
# History:
=============================================================================*/
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#define maxn 205bool mymap[maxn][maxn] = {0};/*记录是否两点相连*/
int link[maxn] = {0};/*记录右边的点所连接的点,没有时置-1*/
bool cover[maxn] = {0};/*记录右边的某个点有没有被覆盖,防止死循环*/
int n = 0, m = 0;bool find (int x)/*寻找左边的点*/
{for (int i = 0; i < m; ++i)/*对右边的点进行遍历*/{if (!cover[i] && mymap[x][i])/*如果该右点没有被覆盖,并且与左点x之间有线相连*/{cover[i] = 1;if (link[i] == -1 || find(link[i])){link[i] = x;return 1;}}}return 0;
}void init()
{memset(mymap, 0, sizeof(mymap));memset(cover, 0, sizeof(cover));memset(link, 0xff, sizeof(link));return;
}int main()
{int link_num = 0;int stall = 0;while (scanf("%d%d", &n, &m) != EOF){int ans = 0;init();for (int i = 0; i < n; ++i){scanf("%d", &link_num);for (int j = 0; j < link_num; ++j){scanf("%d", &stall);mymap[i][stall - 1] = 1;}}for (int i = 0; i < n; ++i)/*遍历所有左点*/{memset(cover, 0, sizeof(cover));if (find(i))++ans;}printf("%d\n", ans);}return 0;
}
第一道二分图,匈牙利算法
![SCOM Rule 介绍 [SCOM中文系列之六]](https://images.cnblogs.com/cnblogs_com/OpsMgr/201103/201103091657084389.png)
