HDOJ 1530 Maximum Clique（最大团模板题）

Maximum Clique

Time Limit: 20000/10000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4365    Accepted Submission(s): 2313

Problem Description

Given a graph G(V, E), aclique is a sub-graph g(v, e), so that for all vertex pairs v1, v2 in v, thereexists an edge (v1, v2) in e. Maximum clique is the clique that has maximumnumber 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 existsbetween 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, thenumber 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

裸的求最大团模板题

先贴出求最大团的优化后的模板

int link[maxn][maxn]; //各点间的连接关系int dp[maxn];         //记录以i出发构成的最大团顶点数量(从i到n-1的最大团)int stk[maxn][maxn];  //第i层中与之相连的第j大的标号int mx;               //最后的结果//一共有n个点，dep表示当前的层数，ns代表与当前层相连并且比ns大的标号个数int dfs(int n,int ns,int dep){    if(ns==0)    {        if(dep>mx)        {            mx=dep;            return 1;        }        return 0;    }    for(int i=0;i<ns;i++)    {        if(ns+dep-i<=mx)            return 0;        int k=stk[dep][i];     //与之相连的第i个点        if(dep+n-k<=mx)        //若当前顶点数量加上还能够增加的最大数量仍小于当前最优解则退出;            return 0;          //若当前顶点数量加上dp[k]仍小于当前最优解则退出;        if(dep+dp[k]<=mx)            return 0;        int cnt=0;        for(int j=i+1;j<ns;j++)        {            int p=stk[dep][j];   //i后边的某个点            if(link[k][p])       //如果i和j相连                stk[dep+1][cnt++]=p;        }        if(dfs(n,cnt,dep+1))            return 1;    }    return 0;}int maxclique(int n){    mx=0;    for(int i=n-1;i>=0;i--)    {        int ns=0;        for(int j=i+1;j<n;j++)        {            if(link[i][j])                stk[1][ns++]=j;        }        dfs(n,ns,1);        dp[i]=mx;    }    return mx;}

题解便是直接套模板。。。

#include <bits/stdc++.h>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define f(i,a,b) for(int i=(a);i<=(b);++i)const int maxn = 105;const int mod = 9973;const int INF = 0x3f3f3f3f;const double eps = 1e-6;#define ll long long#define rush() int T;scanf("%d",&T);while(T--)int link[maxn][maxn];int dp[maxn];      int stk[maxn][maxn]; int mx;            int dfs(int n,int ns,int dep){    if(ns==0)    {        if(dep>mx)        {            mx=dep;            return 1;        }        return 0;    }    for(int i=0;i<ns;i++)    {        if(ns+dep-i<=mx)            return 0;        int k=stk[dep][i];             if(dep+n-k<=mx)                    return 0;                  if(dep+dp[k]<=mx)            return 0;        int cnt=0;        for(int j=i+1;j<ns;j++)        {            int p=stk[dep][j];              if(link[k][p])                      stk[dep+1][cnt++]=p;        }        if(dfs(n,cnt,dep+1))            return 1;    }    return 0;}int maxclique(int n){    mx=0;    for(int i=n-1;i>=0;i--)    {        int ns=0;        for(int j=i+1;j<n;j++)        {            if(link[i][j])                stk[1][ns++]=j;        }        dfs(n,ns,1);        dp[i]=mx;    }    return mx;}int main(){    int n;    while(~scanf("%d",&n)&&n)    {        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)            {                scanf("%d",&link[i][j]);            }        int ans=maxclique(n);        printf("%d\n",ans);    }    return 0;}