hdu 4751 Divide Groups bfs/dfs 连通性的问题 种子染色法

互斥集合，很常见的做法就是把不能在一起的点连上，然后进行染色，如果当前点颜色为1，那么与之相连的点的颜色必须为0。

关键在于判断是否会有一个点有矛盾。

Divide Groups

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1954    Accepted Submission(s): 701

Problem Description

  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.

 

Input

  The input contains several test cases, terminated by EOF.
  Each case starts with a positive integer n (2<=n<=100), which means the number of people joining in the event.
  N lines follow. The i-th line contains some integers which are the id
of students that the i-th student knows, terminated by 0. And the id starts from 1.

 

Output

  If divided successfully, please output "YES" in a line, else output "NO".

 

Sample Input

33 01 01 2 0

 

Sample Output

YES

 

Source

2013 ACM/ICPC Asia Regional Nanjing Online

 

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#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>#include<queue>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=  100  ;int n;bool know[maxn+10][maxn+10];bool G[maxn+10][maxn+10];int color[maxn+10];void build(){    for1(i,n)    {        for1(j,n) if(i!=j&& (!know[i][j] || !know[j][i]) )        {            G[i][j]=1;        }    }}bool bfs(int S){    queue<int >q;    color[S]=0;    q.push(S);    while(!q.empty())    {        int x=q.front();q.pop();        int tc=color[x]^1;        for1(y,n) if(G[x][y])        {            if(~color[y])            {                if(color[y]!=tc) return false;                else continue;            }            color[y]=tc;            q.push(y);        }    }    return true;}bool work(){    memset(color,-1,sizeof color);    for1(i,n) if(color[i]==-1)    {       if(!bfs(i) ) return false;    }    return true;}int main(){    int x;    while(~scanf("%d",&n))    {        memset(know,0,sizeof know);        memset(G,0,sizeof G);        for1(i,n)        {            while(~scanf("%d",&x)&&x)            {                know[i][x]=1;            }        }        build();        puts(work()?"YES":"NO");    }   return 0;}