HDU 5305 Friends (the 6th of the second game of 2015 schools) memory search

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5305

Test Instructions: give you n person, m relationship, the relationship can be online or offline, so that you can ensure that all people online relationship between friends and offline relationship of friends equal situation, such a situation how many kinds.

Ideas: because the online relationship and the number of offline relationships are equal, and M is only 28, so as long as the enumeration of the half of each person's relationship meets the requirements, and according to test instructions M is odd or there is a person's total relationship is odd then there is no situation to meet the requirements, which can be ruled out many cases.

Code:

1#include <cstdio>2#include <cstdlib>3#include <cmath>4#include <cstring>5#include <iostream>6#include <queue>7#include <algorithm>8#include <vector>9 using namespacestd;Ten #defineLL __int64 One  A intf[Ten],on[Ten],off[Ten]; - intM,n; - structnode the { -     intx, y; -}nn[ -]; - voidInit () + { -Memset (F,0,sizeof(f)); +Memset (ON,0,sizeof(on)); Amemset (Off,0,sizeof(off)); at } - intDfsintcot) - { -     intX,y,i,ans; -ans=0; -     if(cot>=m) in     { -          for(i=1; i<=n;i++) to         { +             if(on[i]!=Off[i]) -                 return 0; the         } *         return 1; $     }Panax Notoginsengx=nn[cot].x; -y=nn[cot].y; the     if(on[x]<f[x]/2&&on[y]<f[y]/2) +     { Aon[x]++; theon[y]++; +Ans+=dfs (cot+1); -on[x]--; $on[y]--; $     } -     if(off[x]<f[x]/2&&off[y]<f[y]/2) -     { theoff[x]++; -off[y]++;WuyiAns+=dfs (cot+1); theoff[x]--; -off[y]--; Wu     } -     returnans; About } $  - intMain () - { -     intI,t,ans; A      while(~SCANF ("%d",&T)) +     { the          while(t--) -         { $scanf"%d%d",&n,&m); the init (); the              for(i=0; i<m;i++) the             { thescanf"%d%d",&nn[i].x,&nn[i].y); -f[nn[i].x]++; inf[nn[i].y]++; the             } the             if(m&1) About             { theprintf"0\n"); the                 Continue; the             } +             intflag=0; -              for(i=1; i<=n;i++) the             {Bayi                 if(f[i]&1) the                 { theflag=1; -                      Break; -                 } the             } the             if(flag) the             { theprintf"0\n"); -                 Continue; the             } theAns=dfs (0); theprintf"%d\n", ans);94         } the     } the     return 0; the}

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HDU 5305 Friends (the 6th of the second game of 2015 schools) memory search