bzoj 2530 [Poi2011]Party 构造

2530: [Poi2011]Party

Time Limit: 10 Sec Memory Limit: 128 MBSec Special Judge
Submit: 364 Solved: 213
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Description

给定一张N（保证N是3的倍数）个节点M条边的图，并且保证该图存在一个大小至少为2N/3的团。

请输出该图的任意一个大小为N/3的团。 一个团的定义为节点的一个子集，该子集中的点两两有直接连边。 输入： 第一行是两个整数N,M。 接下来有M行，每行两个整数A,B，表示A和B有连边。保证无重边。 
输出： N/3个整数，表示你找到的团。 数据范围：

3<=N<=3000,[3/2 n(2/3 n -1)]/2<=M<=[n(n-1)/2]

Input

In the first line of the standard input two integers, n and M(3<=N<=3000,[3/2 n(2/3 n -1)]/2<=M<=[n(n-1)/2]), are given, separated by a single space. These denote the number of Byteasar's friends and the number of pairs of his friends who know each other, respectively. Byteasar's friends are numbered from 1 to . Each of the following lines holds two integers separated by a single space. The numbers in line no.i+1(for i=1,2,...,m) are Ai and Bi(1<=Ai<Bi<=N), separated by a single space, which denote that the persons Ai and Bi now each other. Every pair of numbers appears at most once on the input.

Output

In the first and only line of the standard output your program should print N/3numbers, separated by single spaces, in increasing order. These number should specify the numbers of Byteasar's friends whom he should invite to the party. As there are multiple solutions, pick one arbitrarily.

Sample Input

6 10
2 5
1 4
1 5
2 4
1 3
4 5
4 6
3 5
3 4
3 6

Sample Output

2 4

如果两个点没连边，则不可能属于团，所以直接去除，n^2的复杂度即可。

 #include<cstring>

 #include<cmath>

 #include<cstdio>

 #include<algorithm>

 #include<iostream>

 #define N 3007

 #define Wb putchar(' ')

 #define We putchar('\n')

 #define rg register int

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}

     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}

     return x*f;

 }

 inline void write(int x)

 {

     if(x<) putchar('-'),x=-x;

     if (x==) putchar();

     int num=;char c[];

     while(x) c[++num]=(x%)+,x/=;

     while(num) putchar(c[num--]);

 }

 int n,m,cnt;

 int a[N][N],vis[N];

 int main()

 {

     n=read(),m=read();

     for (rg i=;i<=m;i++)

     {

         int x=read(),y=read();

         a[x][y]=a[y][x]=true;

     }

     for (int i=;i<=n;i++)

         for (int j=i+;j<=n&&!vis[i];j++)

             if (!vis[j]&&!a[i][j])

                 vis[i]=vis[j]=true;

     for (int i=;i<=n&&cnt<n/;i++)

         if (!vis[i]) write(i),Wb;

 }

巴特西