Basic knowledge:

Maximum independent set: The vertex set V takes k vertices, 22 of which are not connected to each other.

Maximal Regiment: The vertex set V takes k vertices, and its 22 have edge connections.

Maximum independent set = maximum group of complement graphs. (Fill graph = Full picture-original)

A very detailed explanation: http://www.cnblogs.com/yefeng1627/archive/2013/03/31/2991592.html

Ideas:

Said.. The maximum independent set of the original image is NP problem, then the maximum number of independent concentrated vertices can be obtained by finding the maximum number of vertices in the largest group in the map.

Attach the optimized template:

1#include <cstdio>2#include <cstring>3 #defineN 10104 BOOLFlag[n], a[n][n];5 intans, cnt[n], group[n], N, vis[n];6 //Maximum Regiment: v to take k vertices, two points connected to each other7 //Maximum independent set: K vertices in V, not connected between two points8 //Maximum number of groups = maximum number of independent sets in a complement graph9 Ten BOOLDfsintUintPOS) { One intI, J; A for(i = u+1; I <= N; i++){ - if(Cnt[i]+pos <= ans)return 0; - if(A[u][i]) { the //compare with current group elements, take non-n (i) - for(j =0; J < Pos; J + +)if(!a[i][Vis[j]) Break; - if(j = pos) {//if empty, then all are adjacent to I, then I will be added to the largest group -Vis[pos] =i; + if(Dfs (I, pos+1) )return 1; - } + } A } at if(Pos >ans) { - for(i =0; I < POS; i++ ) -Group[i] = vis[i];//Maximum group element -Ans =Pos; - return 1; - } in return 0; - } to voidMaxclique () + { -ans=-1; the for(intI=n;i>0; i--) * { $vis[0]=i;Panax NotoginsengDFS (I,1); -cnt[i]=ans; the } +}

Bron–kerbosch Algorithm

Code June:

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5#include <string>6#include <queue>7#include <algorithm>8#include <map>9#include <iomanip>Ten#include <climits> One#include <string.h> A#include <numeric> -#include <cmath> -#include <stdlib.h> the#include <vector> -#include <stack> -#include <Set> - #defineFor (x, B, E) for (int x=b;x<= (e); x + +) + #defineREP (x, N) for (int x=0;x< (n); x + +) - #defineINF 1e7 + #defineMAXN 100010 A #defineMAXN 1000010 at #defineMod 1000007 - #defineN 1010 - using namespacestd; -typedefLong LongLL; - - in BOOLFlag[n], a[n][n]; - intans, cnt[n], group[n], N, M, Vis[n]; to BOOLDfsintUintPOS) { + intI, J; - for(i = U +1; I <= N; i++){ the if(Cnt[i] + pos <= ans)return 0; * if(A[u][i]) { $ //compare with current group elements, take non-n (i)Panax Notoginseng for(j =0; J < Pos; J + +)if(!a[i][vis[j]]) Break; - if(j = pos) {//if empty, then all are adjacent to I, then I will be added to the largest group theVis[pos] =i; + if(Dfs (I, POS +1))return 1; A } the } + } - if(Pos >ans) { $ for(i =0; I < POS; i++) $Group[i] = vis[i];//Maximum group element -Ans =Pos; - return 1; the } - return 0;Wuyi } the - voidMaxclique () Wu { -Ans =-1; About for(inti = n; i >0; i--) $ { -vis[0] =i; -DFS (I,1); -Cnt[i] =ans; A } + } the - intMain () { $ intT; thescanf"%d", &T); the while(t--){ thescanf"%d%d", &n, &m); the intx, y; -Memset (A,0,sizeof(a)); in for(inti =0; I < m; i++){ thescanf"%d%d", &x, &y); theA[x][y] = a[y][x] =1; About } the for(inti =1; I <= N; i++) the for(intj =1; J <= N; J + +) the if(i = = j) A[i][j] =0; + ElseA[I][J] ^=1; - Maxclique (); the Bayi if(Ans <0) ans =0; theprintf"%d\n", ans); the for(inti =0; i < ans; i++) -printf (i = =0?"%d":"%d", Group[i]); - if(Ans >0) puts (""); the } the return 0; the}

code June

POJ 1419 (maximum independent set)