HDU 1068 Girls and Boys

English is not very good, but test instructions is probably given a sequence, each of which gives the possibility of being a couple, and now asks you to find a maximum set of people who cannot be lovers.

The problem-solving idea is the maximal independent set, the maximal independent set = vertex number-Maximum match number;

Since the problem is splitting itself, or dividing a point into 2 parts, the final maximum number of matches is divided by 2;

Be sure to remember the initialization, forget the map array initialization = =, timeout several times;

Give an explanation of the URL: http://dsqiu.iteye.com/blog/1689505

1#include <iostream>2#include <string.h>3 using namespacestd;4 #defineMAXN 20055 intLINKER[MAXN];6 BOOLMAP[MAXN][MAXN];7 intun,vn;8 BOOLVISIT[MAXN];9 BOOLDfsintU//find the augmented path from the leftTen { One     intv; A      for(v=0; v<vn;v++)//This vertex number starts at 0 and needs to be modified starting from 1 -         if(map[u][v]&&!Visit[v]) -         { thevisit[v]=true; -             if(linker[v]==-1||DFS (Linker[v])) -{//find an augmented path, reverse -linker[v]=u; +                 return true; -             } +         } A         return false;//don't forget it, always forget it. at } - intHungary () - { -     intres=0; -     intu; -memset (linker,-1,sizeof(linker)); in      for(u=0; u<un;u++) -     { tomemset (Visit,false,sizeof(visit)); +         if(Dfs (U)) res++; -     } the     returnRes; * } $ intMain ()Panax Notoginseng { -     //freopen ("1068.txt", "R", stdin); the     intn,u,v,t; +      while(~SCANF ("%d",&N)) A     { theUN = VN =N; +memset (Map,false,sizeof(map)); -      for(inti =0; i < n; i++) $     { $scanf"%d: (%d)",&u,&t); -           while(t--) -         { thescanf"%d",&v); -MAP[U][V] =true;Wuyi         } the     } -printf"%d\n", N-hungary ()/2); Wu     } -      About     return 0; $}

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HDU 1068 Girls and Boys