HDU 3605 Escape（最大流+合并点）
Escape
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 10116 Accepted Submission(s): 2424
Problem Description
2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.
Input
More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000
Output
Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.
If you can output YES, otherwise output NO.
Sample Input
1 1 1 1 2 2 1 0 1 0 1 1
Sample Output
YES NO
Source
2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU
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题目大意：
有N(1= 解题思路： 题目本身只是一道裸的二分图多重匹配，不过有一点麻烦的是N有点大，可能会导致超时，这样我们就要考虑合并点了。根据Edelweiss写的三条合并规则中的第二条，我们可以发现这些人的边只会指向10个点，那么一共只有2^10=1024种情况。所以我们就可以把相同去向相同的边合并到一起，这样就可以保证一定不会超时了。 AC代码：
#include