poj 3155 Hard Life(最大密度子图，01分数规划)

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http://poj.org/problem?id=3155

大致题意：给出一个无向图，求出它的一个最大密度子图，最大密度子图定义为子图的边数与顶点数的比值。

详见amber论文中关于最大密度子图

本题要注意的地方：

当m为0时也要输出内容。

二分的边界是 1/n/n(high - low >1/n/n) ,这在amber论文引理4.1中有讲解

因为h函数的特性，恒有h >=0，即h函数不是一个严格递减的函数，当减小到0时便不再减小。那么我们要取得的是第一个为0的，所以要二分下界，而不是直接取mid,因为没有意识到函数特性，WA了N次。

二分完毕后，并不能求出正确的low值，ms也是因为该函数的特性，我们还要再求一遍最大流。

根据残量网络求最大密度子图：从源点dfs,走残量网络中流量大于0的边并标记。

#include <stdio.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#define LL long long#define _LL __int64#define eps 1e-8using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 110;const int maxm = 1010;struct node{int u,v;double w;int next,rev;}p[maxm],edge[maxm*6];int s,t,n,m,nn;int cnt,head[maxn];int d[maxn];int dist[maxn],vis[maxn];int ans;void init(){cnt = 0;memset(head,-1,sizeof(head));}void add(int u, int v, double w){edge[cnt] = (struct node){u,v,w,head[u],cnt+1};head[u] = cnt++;edge[cnt] = (struct node){v,u,0,head[v],cnt-1};head[v] = cnt++;}void build(double mid){init();for(int i = 1; i <= m; i++){add(p[i].u,p[i].v,1.0);add(p[i].v,p[i].u,1.0);}for(int i = 1; i <= nn; i++){add(s,i,m);add(i,t,m*1.0+2*mid-d[i]*1.0);}}bool bfs(){    queue <int> que;    memset(dist, 0, sizeof(dist));    memset(vis, 0, sizeof(vis));    while(!que.empty()) que.pop();    vis[s] = 1;    que.push(s);    while(!que.empty())    {        int u = que.front();        que.pop();        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].v;            if(edge[i].w && !vis[v])            {                que.push(v);                vis[v] = 1;                dist[v] = dist[u]+1;            }        }    }    if(dist[t] == 0)return false;return true;}double dfs(int u, double delta){    if(u == t) return delta;    double ret = 0,tmp;for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(edge[i].w && dist[edge[i].v] == dist[u]+1 && (tmp = dfs(v,min(delta,edge[i].w)))){edge[i].w -= tmp;edge[edge[i].rev].w += tmp;return tmp;}}if(!ret) dist[u] = -1;return ret;}double Dinic(){    double ans = 0,res;    while(bfs())    {       while(res = dfs(s,INF))ans += res;    }    return ans;}void dfs_cut(int u){vis[u] = 1;if(u <= nn) ans++;for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(edge[i].w > 0 && !vis[v]){dfs_cut(v);}}}int main(){while(~scanf("%d %d",&n,&m)){if(m == 0){printf("1\n1\n");continue;}memset(d,0,sizeof(d));for(int i = 1; i <= m; i++){scanf("%d %d",&p[i].u,&p[i].v);d[p[i].u]++;d[p[i].v]++;}s = n+1;t = n+2;nn = n;n = t;double low,high,mid,h;low = 0.0;high = m;while(high - low > 1.0/nn/nn){mid = (high + low)/2;build(mid);h = (m*nn*1.0 - Dinic() )/2;if(h > eps)low = mid;else high = mid;}build(low);Dinic();ans = 0;memset(vis,0,sizeof(vis));dfs_cut(s);printf("%d\n",ans);for(int i = 1; i <= nn; i++){if(vis[i])printf("%d\n",i);}}return 0;}

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