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  • UVA 1364 Knights of the Round Table 双连通分量+二分图判定

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<iostream>
    #include<sstream>
    #include<cmath>
    #include<climits>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    #define pb(a) push(a)
    #define INF 0x1f1f1f1f
    #define lson idx<<1,l,mid
    #define rson idx<<1|1,mid+1,r
    #define PI  3.1415926535898
    template<class T> T min(const T& a,const T& b,const T& c) {
        return min(min(a,b),min(a,c));
    }
    template<class T> T max(const T& a,const T& b,const T& c) {
        return max(max(a,b),max(a,c));
    }
    void debug() {
    #ifdef ONLINE_JUDGE
    #else
    
        freopen("d:\in1.txt","r",stdin);
        freopen("d:\out1.txt","w",stdout);
    #endif
    }
    int getch() {
        int ch;
        while((ch=getchar())!=EOF) {
            if(ch!=' '&&ch!='
    ')return ch;
        }
        return EOF;
    }
    
    struct Edge
    {
        int u,v;
    };
    
    const int maxn=1111;
    
    int pre[maxn],iscut[maxn],bccno[maxn];
    int dfs_clock,bcc_cnt;
    
    vector<int> g[maxn],bcc[maxn];
    
    stack<Edge> s;
    
    int dfs(int u,int fa)
    {
        int lowu = pre[u] = ++dfs_clock;
    
        int child=0;
        for(int i=0;i<g[u].size();i++)
        {
            int v=g[u][i];
            Edge e = (Edge){u,v};
    
            if(!pre[v])
            {
                s.push(e);
                child++;
                int lowv=dfs(v,u);
    
                lowu=min(lowu,lowv);
                if(lowv>=pre[u])
                {
                    bcc_cnt++;
                    bcc[bcc_cnt].clear();
                    iscut[u]=true;
                    while(1)
                    {
                        Edge x=s.top();s.pop();
                        if(bccno[x.u]!=bcc_cnt)
                        {
                            bcc[bcc_cnt].push_back(x.u);
                            bccno[x.u]=bcc_cnt;
                        }
    
                        if(bccno[x.v]!=bcc_cnt)
                        {
                            bcc[bcc_cnt].push_back(x.v);
                            bccno[x.v]=bcc_cnt;
                        }
                        if(x.u==u&&x.v==v)break;
                    }
                }
    
            }else if(pre[v]<=pre[u]&&v!=fa)
            {
                lowu=min(pre[v],lowu);
                s.push(e);
            }
        }
        if(fa<0&&child==1)iscut[u]=0;
        return lowu;
    }
    
    void find_bcc(int n)
    {
        memset(bccno,0,sizeof(bccno));
        memset(iscut,0,sizeof(iscut));
        memset(pre,0,sizeof(pre));
        dfs_clock=bcc_cnt=0;
        for(int i=0;i<n;i++)
            if(!pre[i])dfs(i,-1);
    }
    
    int odd[maxn],color[maxn];
    
    bool bipartite(int u,int b)
    {
        for(int i=0;i<g[u].size();i++)
        {
            int v=g[u][i];
            if(bccno[v]!=b)continue;
            if(color[v]==color[u])return false;
            if(!color[v])
            {
                color[v]=3-color[u];
                if(!bipartite(v,b))return false;
            }
    
        }
        return true;
    }
    
    int A[maxn][maxn];
    
    int main()
    {
        int ca=0,n,m;
        while(scanf("%d%d",&n,&m)!=EOF&&n)
        {
            memset(A,0,sizeof(A));
            for(int i=0;i<n;i++)
                g[i].clear();
            for(int i=0;i<m;i++)
            {
                int u,v;
                scanf("%d%d",&u,&v);
                u--;v--;
                A[u][v]=A[v][u]=1;
            }
    
            for(int i=0;i<n;i++)
            {
                for(int j=i+1;j<n;j++)
                {
                    if(!A[i][j])
                    {
                        g[i].push_back(j);
                        g[j].push_back(i);
                    }
                }
            }
            find_bcc(n);
    
            memset(odd,0,sizeof(odd));
            for(int i=1;i<=bcc_cnt;i++)
            {
                memset(color,0,sizeof(color));
                for(int j=0;j<bcc[i].size();j++)bccno[bcc[i][j]]=i;
                int u=bcc[i][0];
                color[u]=1;
                if(!bipartite(u,i))
                    for(int j=0;j<bcc[i].size();j++)
                        odd[bcc[i][j]]=1;
            }
            int res=0;
            for(int i=0;i<n;i++)
                if(!odd[i])res++;
            printf("%d
    ",res);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/BMan/p/3612598.html
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