POJ 2112 Optimal Milking (floyd ＋　二分　＋　网络流）

ＦＬＯＹＤ预处理出每两点之间的最短距离，最大值最小问题一般用二分查找，使用ｄｉｎｉｃ算法求最大流判断是否可行。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#define LL long long
using namespace std;
const int maxn = 10000 + 10;
const int INF = 100000000;
struct Edge
{
int from,to,cap,flow;
Edge(int u,int v,int c,int w):from(u),to(v),cap(c),flow(w){ }
};
int n , m , s , t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
void init()
{
for(int i=0;i<=n+1;i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap)
{
edges.push_back(Edge(from,to,cap,0));
edges.push_back(Edge(to,from,0,0));
int M = edges.size();
G[from].push_back(M-2);
G[to].push_back(M-1);
}
bool BFS()
{
memset(vis,0,sizeof(vis));
queue<int>Q;
Q.push(s);
d[s] = 0;
vis[s] = 1;
while(!Q.empty())
{
int x = Q.front();Q.pop();
for(int i=0;i<G[x].size();i++)
{
Edge& e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a)
{
if(x == t || a == 0) return a;
int flow = 0 , f;
for(int &i = cur[x];i<G[x].size();i++)
{
Edge& e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = DFS(e.to,min(a,e.cap-e.flow))) > 0)
{
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int Maxflow()
{
int flow = 0;
while(BFS())
{
memset(cur,0,sizeof(cur));
flow += DFS(s,INF);
}
return flow;
}
int K , C , M;
int dis[500][500];
void creat_graph(int limit)
{
init();
for(int i=K+1;i<=n;i++) AddEdge(0,i,1);
for(int i=1;i<=K;i++) AddEdge(i,n+1,M);
for(int i=K+1;i<=n;i++)
{
for(int j=1;j<=K;j++)
{
if(dis[i][j] <= limit) AddEdge(i,j,1);
}
}
s = 0 , t = n+1;
}
int main()
{
while(scanf("%d%d%d",&K,&C,&M)!=EOF)
{
n = K + C;
memset(dis,0,sizeof(dis));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&dis[i][j]);
if(dis[i][j] == 0) dis[i][j] = INF;
}
}
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]);
}
}
int L = 0 , R = 10000;
int ans;
while(L < R)
{
int m = (L + R) / 2;
ans = 0;
creat_graph(m);
ans = Maxflow();
if(ans >= C) R = m;
else L = m + 1;
}
printf("%d\n",R);
}
return 0;
}