poj 2112最佳挤奶方案

用FLOYD求出任意两点最小距离，用Dinic求最大流，用二分法搜索最大距离最小值#include <iostream>#include <cstring>#include <cstdio>using namespace std;#define MAX 300#define INF 1000000int dis[MAX][MAX];int map[MAX][MAX];bool sign[MAX][MAX];bool used[MAX];int K,C,n,M;int min(int a,int b){  return a < b ? a : b;}void Build_Graph(int min_max){  int i,j;  memset(map,0,sizeof(map));  for(i = K+1; i <=n; i++)   //从源点向每头奶牛建一条边，容量为1    map[0][i] = 1;  for(i = 1; i <= K; i++)    //从取奶器向汇点建边，容量为M    map[i][n+1] = M;  for(i = K+1;i <= n;i++) //从每头奶牛向取奶器建边，容量为1    {      for(j = 1; j <= K;j++){  if(dis[i][j] <= min_max)    map[i][j] = 1;}    }}bool BFS()  //构建层次网络{  memset(used,0,sizeof(used));  memset(sign,0,sizeof(sign));  int queue[100*MAX] = {0};  queue[0] = 0;  used[0] = 1;  int t = 1,f = 0;  while(f < t)    {      for(int i = 0; i <= n + 1; i++){  if(!used[i] && map[queue[f]][i])    {      queue[t++] = i;      used[i] = 1;      sign[queue[f]][i] = 1;    }}      f++;    }  if(used[n+1])    return true;  else    return false;}int DFS(int v,int sum)  //DFS增广{  int i,s,t;  if(v == n + 1)    return sum;  s = sum;  for(i = 0; i <= n+1;i++)    {      if(sign[v][i]){  t = DFS(i,min(map[v][i],sum));  map[v][i] -= t;  map[i][v] += t;  sum -= t;}    }  return s-sum;}int main(){  int i,j,k,L,R,mid,ans;  cin>>K>>C>>M;  n = K + C;  for(i = 1; i <= n; i++)    {      for(j = 1; j <= n; j++){  cin>>dis[i][j];  if(dis[i][j] == 0)    dis[i][j] = INF;}    }  for(k = 1; k <= n; k++)  //floyd求最短距离    for(i = 1; i <= n; i++)      for(j = 1; j <= n; j++){  dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]);}  L = 0, R = 10000;  while(L < R)  //二分搜索    {      mid = (L + R)/2;      ans = 0;      Build_Graph(mid);      while(BFS()) ans += DFS(0,INF);   //dinic求最大流      if(ans >= C) R = mid;      else L = mid + 1;    }  cout<<R<<endl;  return 0;}