bzoj 1312 最大密度子图

晕，m=0是要输出1（弄的我还找管理员要数据，但明显题意是叫我们输出0呀）

最大密度子图，把边转换成点，然后二分答案，跑最大权闭合子图判定是否可行。

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <algorithm>
 #define N 1110
 #define oo 0x3f3f3f3f
 using namespace std;

 struct Edge {
     int u, v, f;
     Edge( int u, int v, int f ):u(u),v(v),f(f){}
 };
 int gcd( int a, int b ) {
     return b?gcd(b,a%b):a;
 }
 struct Pair {
     int a, b;
     Pair( int aa, int bb ) {
         int cd = gcd(aa,bb);
         a=aa/cd;
         b=bb/cd;
     }
     bool operator<( const Pair &o ) const {
         return a*o.b < o.a*b;
     }
     bool operator==( const Pair &o ) const {
         return a*o.b==o.a*b;
     }
 };
 int n, m;
 vector<Edge> edge;
 vector<int> g[N];
 vector<Pair> prs;
 int dep[N], cur[N], qu[N], bg, ed;
 int uu[N], vv[N], idx[N], src, dst, idc;
 bool vis[N];

 void init() {
     for( int i=; i<=dst; i++ )
         g[i].clear();
     edge.clear();
 }
 void adde( int u, int v, int f ) {
     g[u].push_back( edge.size() );
     edge.push_back( Edge(u,v,f) );
     g[v].push_back( edge.size() );
     edge.push_back( Edge(v,u,) );
 }
 bool bfs() {
     memset( dep, , sizeof(dep) );
     qu[bg=ed=] = src;
     dep[src] = ;
     while( bg<=ed ) {
         int u=qu[bg++];
         for( int t=; t<g[u].size(); t++ ) {
             Edge &e=edge[g[u][t]];
             if( e.f && !dep[e.v] ) {
                 dep[e.v]=dep[e.u]+;
                 qu[++ed] = e.v;
             }
         }
     }
     return dep[dst];
 }
 int dfs( int u, int a ) {
     if( u==dst || a== ) return a;
     int remain=a, past=, na;
     for( int &t=cur[u]; t<g[u].size(); t++ ) {
         Edge &e=edge[g[u][t]];
         Edge &ve=edge[g[u][t]^];
         if( e.f && dep[e.v]==dep[e.u]+ && (na=dfs(e.v,min(e.f,remain))) ) {
             remain -= na;
             past += na;
             e.f -= na;
             ve.f += na;
             if( !remain ) break;
         }
     }
     return past;
 }
 int maxflow() {
     int flow = ;
     while( bfs() ) {
         memset( cur, , sizeof(cur) );
         flow += dfs(src,oo);
     }
     return flow;
 }
 void makeid() {
     src = n+m+;
     dst = src+;
     idc = n;
     for( int i=; i<=m; i++ ) idx[i] = ++idc;
 }
 void rebuild( int a, int b ) {
     init();
     for( int i=; i<=m; i++ ) {
         adde( src, idx[i], b );
         adde( idx[i], uu[i], oo );
         adde( idx[i], vv[i], oo );
     }
     for( int i=; i<=n; i++ )
         adde( i, dst, a );
 }
 bool ok( int a, int b ) {
     rebuild(a,b);
     return m*b-maxflow() > ;
 }
 int calc() {
     int cnt = ;
     qu[bg=ed=] = dst;
     vis[dst] = true;
     while( bg<=ed ) {
         int u=qu[bg++];
         for( int t=; t<g[u].size(); t++ ) {
             Edge &e = edge[g[u][t]^];
             if( e.f && !vis[e.u] ) {
                 vis[e.u] = true;
                 cnt += <=e.u&&e.u<=n;
                 qu[++ed] = e.u;
             }
         }
     }
     return n-cnt;
 }
 int binary() {
     for( int a=; a<=m; a++ )
         for( int b=; b<=n; b++ )
             prs.push_back( Pair(a,b) );
     sort( prs.begin(), prs.end() );
     prs.erase( unique(prs.begin(),prs.end()), prs.end() );
     int lf=, rg=prs.size()-;
     while( lf<rg ) {
         int mid = (lf+rg)>>;
         if( ok(prs[mid].a,prs[mid].b) )
             lf=mid+;
         else
             rg=mid;
     }
     ok(prs[lf].a,prs[lf].b);
     return calc();
 }
 int main() {
     scanf( "%d%d", &n, &m );
     if( m== ) {
         printf( "1\n" );
         return ;
     }
     for( int i=; i<=m; i++ )
         scanf( "%d%d", uu+i, vv+i );
     makeid();
     printf( "%d\n", binary() );
 }