poj 2112 Optimal Milking(二分+最大流)

来源：互联网发布：vhf天线仿真软件编辑：程序博客网时间：2024/05/16 06:08

题意：有K个挤奶器和C头奶牛，没台挤奶器最多能让M头奶牛挤奶，奶牛到挤奶器需要走一段路程，安排每头奶牛到一台挤奶器挤奶，令所有奶牛中要走的最大距离最小。

思路：比较明显的最大流，最大距离最小只要二分就行了。另外，可以先用floyd处理下两点之间的最短路……

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<set>#include<stack>#include<cmath>#include<vector>#define inf 0x3f3f3f3f#define Inf 0x3FFFFFFFFFFFFFFFLL#define eps 1e-9#define pi acos(-1.0)using namespace std;typedef long long ll;const int maxn=200+55;const int maxm=10000+10;int dis[maxn][maxn],d[maxn],cur[maxn];int K,C,M,N,S,T;int head[maxn],nEdge;bool vis[maxn];struct Edge{    int to,next,cap,flow;    Edge (){};    Edge (int v,int nx,int ca,int fl){to=v;next=nx;cap=ca;flow=fl;}}edges[maxm<<1];void AddEdge(int from,int to,int cap){    nEdge++;    edges[nEdge]=Edge(to,head[from],cap,0);    head[from]=nEdge++;    edges[nEdge]=Edge(from,head[to],0,0);    head[to]=nEdge;}void floyd(){    for(int k=1;k<=N;++k)        for(int i=1;i<=N;++i)            for(int j=1;j<=N;++j)                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);}bool BFS(){    memset(vis,0,sizeof(vis));    memset(d,0,sizeof(d));    d[S]=0;vis[S]=true;    queue<int>q;    q.push(S);    while(!q.empty())    {        int u=q.front();q.pop();        for(int k=head[u];k!=-1;k=edges[k].next)        {            Edge e=edges[k];            if(!vis[e.to]&&e.cap>e.flow)            {                d[e.to]=d[u]+1;                vis[e.to]=true;                q.push(e.to);            }        }    }    return vis[T];}int DFS(int u,int a){    if(u==T||a==0) return a;    int flow=0,f;    for(int &k=cur[u];k!=-1;k=edges[k].next)    {        Edge e=edges[k];        if(d[e.to]==d[u]+1&&(f=DFS(e.to,min(e.cap-e.flow,a)))>0)        {            edges[k].flow+=f;            edges[k^1].flow-=f;            flow+=f;            a-=f;            if(a==0) break;        }    }    return flow;}int Maxflow(){    int flow=0;    while(BFS())    {        for(int i=0;i<=T;++i) cur[i]=head[i];        flow+=DFS(S,inf);    }    return flow;}void buildGragh(int lim){    nEdge=-1;    memset(head,0xff,sizeof(head));    S=0,T=N+1;    for(int i=1;i<=C;++i)    {        AddEdge(S,K+i,1);        for(int j=1;j<=K;++j)           if(dis[K+i][j]<=lim) AddEdge(K+i,j,1);    }    for(int i=1;i<=K;++i)        AddEdge(i,T,M);}int solve(){    int L=0,R=inf;    int val;    while(L<R)    {        int m=(L+R)>>1;        buildGragh(m);        val=Maxflow();        if(val==C)            R=m;        else L=m+1;    }    return R;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    scanf("%d%d%d",&K,&C,&M);    N=C+K;    for(int i=1;i<=N;++i)      for(int j=1;j<=N;++j)      {          scanf("%d",&dis[i][j]);          if(dis[i][j]==0&&i!=j) dis[i][j]=inf;      }    floyd();    int ans=solve();    printf("%d\n",ans);    return 0;}

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