poj 1419 Graph Coloring(无向图的最大独立集)
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Graph Coloring
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3645 Accepted: 1612 Special Judge
Description
You are to write a program that tries to find an optimal coloring for a given graph. Colors are applied to the nodes of the graph and the only available colors are black and white. The coloring of the graph is called optimal if a maximum of nodes is black. The coloring is restricted by the rule that no two connected nodes may be black.
Figure 1: An optimal graph with three black nodes
Figure 1: An optimal graph with three black nodes
Input
The graph is given as a set of nodes denoted by numbers 1...n, n <= 100, and a set of undirected edges denoted by pairs of node numbers (n1, n2), n1 != n2. The input file contains m graphs. The number m is given on the first line. The first line of each graph contains n and k, the number of nodes and the number of edges, respectively. The following k lines contain the edges given by a pair of node numbers, which are separated by a space.
Output
The output should consists of 2m lines, two lines for each graph found in the input file. The first line of should contain the maximum number of nodes that can be colored black in the graph. The second line should contain one possible optimal coloring. It is given by the list of black nodes, separated by a blank.
Sample Input
16 81 21 32 42 53 43 64 65 6
Sample Output
31 4 5
题意:给出一个无向图,要求给图中的点着色,可以用白色或黑色,规定不能有两个黑色的点相邻,问最多可以给多少个点着黑色。
思路:求无向图的最大独立集。最大独立集=补图的最大团。
AC代码:
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <map>#include <cstdlib>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define ll long long#define eps 1e-6using namespace std;const int maxn=105;bool G[maxn][maxn];bool vis[maxn],color[maxn];int n,m,cnt,bestn;void dfs(int u){ if(u>n) { if(cnt>bestn) { bestn=cnt; for(int i=1;i<=n;i++) color[i]=vis[i]; } return; } bool ok=true; for(int i=1;i<u;i++) if(vis[i]&&!G[i][u]) { ok=false; break; } if(ok) { vis[u]=true; cnt++; dfs(u+1); vis[u]=false; cnt--; } if(cnt+n-u>bestn) dfs(u+1);}int main(){ int t,a,b; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) for(int j=i;j<=n;j++) G[i][j]=G[j][i]=true; while(m--) { scanf("%d%d",&a,&b); G[a][b]=G[b][a]=false; } cnt=bestn=0; memset(vis,false,sizeof(vis)); dfs(1); printf("%d\n",bestn); for(int i=1;i<=n;i++) if(color[i]) printf("%d ",i); puts(""); } return 0;}
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