poj 2112 Optimal Milking(最大費用流+dinic算法+二分搜索)


Optimal Milking
Time Limit: 2000MS   Memory Limit: 30000K
Total Submissions: 9253   Accepted: 3341
Case Time Limit: 1000MS

Description

FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C. 

Each milking point can "process" at most M (1 <= M <= 15) cows each day. 

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse several paths on the way to their milking machine. 

Input

* Line 1: A single line with three space-separated integers: K, C, and M. 

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line. 

Output

A single line with a single integer that is the minimum possible total distance for the furthest walking cow. 

Sample Input

2 3 2
0 3 2 1 1
3 0 3 2 0
2 3 0 1 0
1 2 1 0 2
1 0 0 2 0

Sample Output

2

Source


思路:這題,得用二分步長,判流量。用二維數組的dinic很險。很容易TLE。


#include<iostream>
#include<cstring>
using namespace std;
const int mm=310;
const int oo=1e9;
int dis[mm][mm],le[mm][mm],g[mm][mm],pre[mm],q[mm];
int k,c,m,z;
void build(int x)///初始化數據,即是加邊
{ memset(g,0,sizeof(g));
  for(int i=1;i<=k;i++)
    g[0][i]=m;
  for(int i=k+1;i<=z;i++)
    g[i][z+1]=1;
  for(int i=1;i<=k;i++)
    for(int j=k+1;j<=z;j++)
    if(dis[i][j]<=x)
    g[i][j]=1;
}
int bfs()///廣搜找殘留網絡
{
  int qs=0,qt=1;
  memset(pre,-1,sizeof(pre));
  memset(le,0,sizeof(le));
  q[qs]=0;pre[0]=-2;int zz;
  while(qs!=qt&&pre[z+1]==-1)
  {
    zz=q[qs++];qs%=mm;
    for(int i=0;i<=z+1;i++)
    {
      if(pre[i]==-1&&g[zz][i])
      { le[zz][i]=1;
        pre[i]=zz;q[qt++]=i;qt%=mm;
      }
    }
  }
  if(pre[z+1]==-1)return 0;
  else return 1;
}
int _min(int x,int y)
{
  if(x<y)return x;
  return y;
}
int dfs(int v,int sum)///深搜遍歷所有增廣路徑
{ int ret=sum,rt;
  if(v==z+1)return sum;
  for(int i=0;i<=z+1;i++)
  {
    if(le[v][i])
    {
      le[v][i]=0;
      rt=dfs(i,_min(sum,g[v][i]));
      g[v][i]-=rt;g[i][v]+=rt;
      sum-=rt;
    }
  }
  return ret-sum;
}
int main()
{
  while(cin>>k>>c>>m)
  {
    z=k+c;
    for(int i=1;i<=z;i++)
      for(int j=1;j<=z;j++)
    {
      cin>>dis[i][j];if(!dis[i][j])dis[i][j]=oo;
    }
    for(int l=1;l<=z;l++)
      for(int i=1;i<=z;i++)
      for(int j=1;j<=z;j++)
      if(dis[i][l]^oo&&dis[l][j]^oo)
    {
      int zz=dis[i][l]+dis[l][j];
      dis[i][j]=dis[i][j]<zz?dis[i][j]:zz;
    }
    int l=0,r=10009,mid,ret;
    while(l<r)///二分距離
    {
      ret=0;
      mid=(l+r)/2;
      build(mid);ret=0;
      while(bfs())
        {
         ret+=dfs(0,oo);
        }
      if(ret>=c)r=mid;
      else l=mid+1;
     }
    cout<<r<<"\n";
  }
}




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