[poj 2112]Optimal Milking

题目大意：有K (1 <= K <= 30)台挤奶机，C (1 <= C <= 200)头奶牛。每头牛到每台奶牛机之间有多条路。每台挤奶机每天最多处理M (1 <= M <= 15) 头奶牛。求一个分配方案，使得在处理全部奶牛的情况下，距离奶牛机最远的奶牛距离最短。

网络流裸题。先floyd预处理出每头奶牛到每台挤奶机的距离，然后二分这个最远距离（显然最大值最小很容易想到二分）。每一对距离<=mid的奶牛与挤奶机就可以匹配。之后从S到每个奶牛连一条流量为1的边，代表每个奶牛都得被挤奶；然后从每个奶牛到每个可以匹配到的挤奶机连一条流量为1的边；最后从每个挤奶机到T连一条流量为M的边,代表每台挤奶机最多处理M头奶牛。跑一遍最大流，判断一下流量是否等于C就行了。

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <algorithm>#include <vector>#include <deque>#include <queue>#include <map>#include <set>#include <ctime>using namespace std;const int MAXN=35,MAXM=205,INF=(1<<29);int n,m,K,S,T,tot,fst[MAXN+MAXM],pre[MAXN*MAXM*2],to[MAXN*MAXM*2],cap[MAXN*MAXM*2],dis[MAXN+MAXM][MAXN+MAXM],head,tail,Q[MAXN+MAXM],depth[MAXN+MAXM]; bool vis[MAXN+MAXM]; int Get(){     char ch; int v=0; bool f=false;    while (!isdigit(ch=getchar())) if (ch=='-') f=true; v=ch-48;    while (isdigit(ch=getchar())) v=v*10+ch-48;     if (f) return -v;else return v;   }void add(int x,int y,int c){     pre[++tot]=fst[x],fst[x]=tot,to[tot]=y,cap[tot]=c;    pre[++tot]=fst[y],fst[y]=tot,to[tot]=x,cap[tot]=0; }bool Bfs(){     for (int i=S;i<=T;i++) vis[i]=false,depth[i]=0;    head=tail=1; Q[tail++]=S,vis[S]=true,depth[S]=1;     while (head!=tail)      {         int x=Q[head++];         for (int i=fst[x];i;i=pre[i])          {             int y=to[i];            if (!vis[y] && cap[i]>0) depth[y]=depth[x]+1,Q[tail++]=y,vis[y]=true;           }      }    return vis[T];}int dfs(int x,int flow){       if (x==T || flow==0) return flow;    int f,nowflow=0;    for (int i=fst[x];i;i=pre[i])      {         int y=to[i];        if (cap[i]>0 && depth[y]==depth[x]+1)          {             f=dfs(y,min(flow,cap[i])); if (f==0) continue;            nowflow+=f,flow-=f,cap[i]-=f,cap[i^1]+=f;           }      }    if (nowflow==0) depth[x]=0;    return nowflow; }int Dinic() { int res=0; while (Bfs()) res+=dfs(S,INF); return res; }bool check(int maxdis){       S=0,T=n+m+1; tot=1;     for (int i=S;i<=T;i++) fst[i]=0;    for (int i=1;i<=n;i++) add(i,T,K);    for (int i=1;i<=m;i++) add(S,i+n,1);     for (int i=1;i<=m;i++) for (int j=1;j<=n;j++) if (dis[i+n][j]<=maxdis) add(i+n,j,1);    return (Dinic()==m); }void work(){     int L=0,R=(MAXN+MAXM)*MAXM,mid;    while (L<R)      {         mid=(L+R)>>1;         if (check(mid)) R=mid;else L=mid+1;      }    printf("%d\n",L);}int main(){     //freopen("milk.in","r",stdin);    //freopen("milk.out","w",stdout);    n=Get(),m=Get(),K=Get(); int d;    for (int i=1;i<=n+m;i++)       for (int j=1;j<=n+m;j++)         {           d=Get();          if (d==0) dis[i][j]=INF;else dis[i][j]=d;        }    for (int k=1;k<=n+m;k++)      for (int i=1;i<=n+m;i++)         for (int j=1;j<=n+m;j++) dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);    work();    return 0;}