Maximum Clique

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4546    Accepted Submission(s): 2392 Problem Description Given a graph G(V, E), a clique is a sub-graph g(v, e), so that for all vertex pairs v1, v2 in v, there exists an edge (v1, v2) in e. Maximum clique is the clique that has maximum number of vertex.   Input Input contains multiple tests. For each test: The first line has one integer n, the number of vertex. (1 < n <= 50) The following n lines has n 0 or 1 each, indicating whether an edge exists between i (line number) and j (column number). A test with n = 0 signals the end of input. This test should not be processed.   Output One number for each test, the number of vertex in maximum clique.   Sample Input 5 0 1 1 0 1 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 0 0   Sample Output 4   Author CHENG, Long   Source ZOJ Monthly, February 2003 题意：找出图的最大团，最大团即点数最多的完全图，完全图即两两点都有边相连。

思路：dfs。

稍微暴力版：

# include <iostream> # include <cstdio> # include <cstring> using namespace std; int m[53][53], n, vis[53], cnt, ans; void dfs(int x) { if(x > n) { ans = max(ans, cnt); return; } int flag = 1; for(int i=1; i<x; ++i) { if(vis[i] && !m[i][x]) { flag = 0; break; } } if(flag) { vis[x] = 1, ++cnt; dfs(x+1); vis[x] = 0, --cnt; } if(n-x+cnt > ans) dfs(x+1); } int main() { while(~scanf("%d",&n),n) { cnt = ans = 0; memset(vis, 0, sizeof(vis)); for(int i=1; i<=n; ++i) for(int j=1; j<=n; ++j) scanf("%d",&m[i][j]); dfs(1); printf("%d\n",ans); } return 0; } dp版：

# include <iostream> # include <cstdio> # include <cstring> using namespace std; int m[53][53], n, ans, dp[53], a[53][53];//dp[i]表示在第i个数到后面可以选的最大团。 void dfs(int up, int dep) { if(up == 0) { ans = max(ans, dep); return; } for(int i=0; i<up; ++i) { int cnt = 0; int u = a[dep][i]; if(dep + n - u + 1 <= ans) return; if(dep + dp[u] <= ans) return; for(int j=i+1; j<up; ++j) { int v = a[dep][j]; if(m[u][v]) a[dep+1][cnt++] = v; } dfs(cnt, dep+1); } } int clique() { for(int i=n; i>0; --i) { int cnt = 0; for(int j=i+1; j<=n; ++j) { if(m[i][j]) a[1][cnt++] = j; } dfs(cnt, 1); dp[i] = ans; } return ans; } int main() { while(~scanf("%d",&n),n) { ans = 0; memset(a, 0, sizeof(a)); memset(dp, 0, sizeof(dp)); for(int i=1; i<=n; ++i) for(int j=1; j<=n; ++j) scanf("%d",&m[i][j]); printf("%d\n",clique()); } return 0; }