hdu4751Divide Groups【判斷二分圖】

Total Submission(s): 1904    Accepted Submission(s): 676

  This year is the 60th anniversary of NJUST, and to make the celebration more colorful, Tom200 is going to invite distinguished alumnus back to visit and take photos.
  After carefully planning, Tom200 announced his activity plan, one that contains two characters:
  1. Whether the effect of the event are good or bad has nothing to do with the number of people join in.
  2. The more people joining in one activity know each other, the more interesting the activity will be. Therefore, the best state is that, everyone knows each other.
  The event appeals to a great number of alumnus, and Tom200 finds that they may not know each other or may just unilaterally recognize others. To improve the activities effects, Tom200 has to divide all those who signed up into groups to take part in the activity at different time. As we know, one's energy is limited, and Tom200 can hold activity twice. Tom200 already knows the relationship of each two person, but he cannot divide them because the number is too large.
  Now Tom200 turns to you for help. Given the information, can you tell if it is possible to complete the dividing mission to make the two activity in best state.

自己好菜啊QAQ 題意：給出彼此認識的關係（不傳遞），問是否可以分成兩部分使得兩部分中都互相認識

裸的判斷二分圖，模板手誤寫錯了兩個單詞，最大的問題是最開始為標記陣列賦值的時候是單向的！加邊的時候是雙向的==

再有，難道不應該染色的時候只在for迴圈外面只賦值一次嗎？為什麼寫裡面寫外面都對啊==

#include <iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int col[200],n,x;
vector<int>G[200];
bool flag[200][200];
bool dfs(int u,int color)
{
    col[u]=color;
    for(int i=0;i<G[u].size();i++)
    {
        int v=G[u][i];
        if(col[v]==color) return false;
        if(col[v]==-1)
        {
            if(!dfs(v,!color)) return false;
        }
    }
    return true;
}
int main()
{
  //  freopen("cin.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++) G[i].clear();
        memset(flag,0,sizeof(flag));
        for(int i=1;i<=n;i++)
        {
            while(scanf("%d",&x)&&x)
            {
               flag[i][x]=1;
              // flag[x][i]=1;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j) continue;
                if(0==flag[i][j]||0==flag[j][i])
                {
                    G[i].push_back(j);G[j].push_back(i);
                }
             }
        }

        bool fl=true;
        memset(col,-1,sizeof(col));
        for(int i=1;i<=n&&fl;i++)
        {
             
             if(col[i]==-1)
            {
                if(!dfs(i,0)) fl=false;
            }
        }
        if(fl) puts("YES");
        else puts("NO");
    }
    return 0;
}