[最大密度子图 最小割] BZOJ 1312 Neerc2006 Hard Life

就是裸题啦

比较好的建模推导：http://www.cnblogs.com/vongang/archive/2012/10/25/2740042.html

自己也也要做到能推呢

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#define V G[p].v#define eps 1e-5#define cl(x) memset(x,0,sizeof(x))using namespace std;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }    return *p1++;}inline void read(int &x){    char c=nc(),b=1;    for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;    for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline int dcmp(double a,double b){    if (fabs(a-b)<eps) return 0;    if (a<b) return -1; return 1;}namespace DINIC{    const int N=1005;    const int M=10005;    struct edge{        int u,v; double f;        int next;    }G[M];    int head[N],inum=1;    inline void add(int u,int v,double f,int p){        G[p].u=u; G[p].v=v; G[p].f=f; G[p].next=head[u]; head[u]=p;    }    inline void link(int u,int v,double f){        add(u,v,f,++inum); add(v,u,0,++inum);    }    int S,T;    int dis[N];    int Q[N],l,r;    inline bool bfs(){        memset(dis,-1,sizeof(dis)); l=r=-1;        Q[++r]=S; dis[S]=1;        while (l<r){            int u=Q[++l];            for (int p=head[u];p;p=G[p].next)                if (dcmp(G[p].f,0)>0 && dis[V]==-1){                    dis[V]=dis[u]+1; Q[++r]=V;                    if (V==T) return 1;                }        }        return 0;    }    int cur[N];    double dfs(int u,double flow){        if (u==T) return flow;        double used=0,now;        for (int p=cur[u];p;p=G[p].next)        {            cur[u]=p;            if (dcmp(G[p].f,0)>0 && dis[V]==dis[u]+1)            {                now=dfs(V,min(flow-used,G[p].f));                G[p].f-=now,G[p^1].f+=now;                used+=now; if (dcmp(flow,used)==0) break;            }        }        if (dcmp(used,0)==0) dis[u]=-1;        return used;    }    double Dinic(){        double ret=0;        while (bfs())            memcpy(cur,head,sizeof(cur)),ret+=dfs(S,1e40);        return ret;    }    void clear(){        cl(head); inum=1;    }    int tot;    int vst[N];    void dfs(int u){        vst[u]=1; tot++;        for (int p=head[u];p;p=G[p].next)            if (dcmp(G[p].f,0)>0 && !vst[V])                dfs(V);    }    inline int getans(){        tot=-1; dfs(S);        return tot;    }}int u[1005],v[1005],deg[1005];int n,m;inline bool check(double mid){    using namespace DINIC;    double U=m;    clear(); S=n+1; T=n+2;    for (int i=1;i<=m;i++) link(u[i],v[i],1),link(v[i],u[i],1);    for (int i=1;i<=n;i++) link(S,i,U);    for (int i=1;i<=n;i++) link(i,T,U+2*mid-deg[i]);    double ret=Dinic();    return dcmp(U*n-ret,0)>0;}inline void Solve(){    using namespace DINIC;    double L=0,R=m,MID;    while (R-L>=1.0/n/n)        if (check(MID=(L+R)/2))            L=MID;        else            R=MID;    check(L);    int ans=getans();    printf("%d\n",ans);}int main(){    freopen("t.in","r",stdin);    freopen("t.out","w",stdout);    read(n); read(m);    if (!m) return printf("1\n"),0;    for (int i=1;i<=m;i++)        read(u[i]),read(v[i]),deg[u[i]]++,deg[v[i]]++;    Solve();    return 0;}