pku 2112 Optimal Milking floyd + 二分 + 网络流
http://poj.org/problem?id=2112
题意：
给出K个挤奶的机器，C个奶牛,以及每个挤奶机器每天最多服务M头奶牛，给出他们的之间的距离。求奶牛到挤奶器处产奶，其满足小于等于M的情况下，最远距离的最小值。
思路：
不得不说这是一个很好的题目，首先我们看到的是机器每天工作量的限制，让我们联想到流的限制。我们首先二分枚举一个距离，然后建图源点s到挤奶器建边权值为M，奶牛到汇点e建边权值为1.然后枚举奶牛与挤奶器之间的距离，如果小于mid就建立一条权值为1的边，然后求最大流如果小于C继续增加，否则减小。 这里建图之前先用floyd求出任意两点之间的最短距离，这样保证枚举的任意两点距离最短。
//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#define CL(arr, val) memset(arr, val, sizeof(arr))
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x) (x) << 1
#define R(x) (x) << 1 | 1
#define MID(l, r) (l + r) >> 1
#define Min(x, y) (x) < (y) ? (x) : (y)
#define Max(x, y) (x) < (y) ? (y) : (x)
#define E(x) (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)int
#define OUT(x) printf("%I64d\n", x)
#define lowbit(x) (x)&(-x)
#define Read() freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);
#define N 350
using namespace std;
struct node
{
int v,w;
int next;
}g[N*N];
int head[N],ct;
int mat[N][N];
int level[N];
int q[N*1000];
int K,C,M,n;
void add(int u,int v,int w)
{
g[ct].v = v;
g[ct].w = w;
g[ct].next = head[u];
head[u] = ct++;
g[ct].v = u;
g[ct].w = 0;
g[ct].next = head[v];
head[v] = ct++;
}
bool layer(int s,int e)
{
int i;
CL(level,-1);
queue<int>q;
q.push(s);
level[s] = 1;
while (!q.empty())
{
int u = q.front(); q.pop();
for (i = head[u]; i != -1; i = g[i].next)
{
int v = g[i].v;
int w = g[i].w;
if (w > 0 && level[v] == -1)
{
level[v] = level[u] + 1;
if (v == e) return true;
q.push(v);
}
}
}
return false;
}
int find(int s,int e)
{
int i;
int top = 1,u;
int ans = 0;
while (top)
{
if (top == 1) u = s;
else u = g[q[top - 1]].v;
if (u == e)
{
int MIN = inf,pos = 0;
for (i = 1; i <= top - 1; ++i)
{
if (MIN > g[q[i]].w)
{
MIN = g[q[i]].w;
pos = i;
}
}
for (i = 1; i <= top - 1; ++i)
{
g[q[i]].w -= MIN;
g[q[i]^1].w += MIN;
}
ans += MIN;
top = pos;
}
else
{
for (i = head[u]; i != -1; i = g[i].next)
{
int w = g[i].w;
int v = g[i].v;
if (level[v] == level[u] + 1 && w > 0)
{
q[top++] = i;
break;
}
}
if (i == -1)
{
top--;
level[u] = -1;
}
}
}
return ans;
}
int dinic(int s,int e)
{
int ans = 0;
while (layer(s,e)) ans += find(s,e);
return ans;
}
void build(int s,int e,int mid)
{
int i,j;
CL(head,-1); ct = 0;
for (i = 1; i <= K; ++i) add(s,i,M);
for (i = K + 1; i <= n; ++i) add(i,e,1);
for (i = 1; i <= K; ++i)
{
for (j = K + 1; j <= n; ++j)
{
if (mat[i][j] <= mid) add(i,j,1);
}
}
}
void floyd()
{
int i,j,k;
for (k = 1; k <= n; ++k)
{
for (i = 1; i <= n; ++i)
{
for (j = 1; j <= n; ++j)
{
if (mat[i][k] != inf && mat[k][j] != inf && mat[i][j] > mat[i][k] + mat[k][j])
{
mat[i][j] = mat[i][k] + mat[k][j];
}
}
}
}
}
int main()
{
//Read();
int i,j;
while (~scanf("%d%d%d",&K,&C,&M))
{
n = K + C;
for (i = 1; i <= n; ++i)
{
for (j = 1; j <= n; ++j)
{
scanf("%d",&mat[i][j]);
if (mat[i][j] == 0) mat[i][j] = inf;
}
}
floyd();
int s = 0, e = n + 1;
int l = 0;
int r = 200*n;
int res = 0;
while (l <= r)
{
int mid = (l + r)/2;
build(s,e,mid);
int ans = dinic(s,e);
// printf("%d %d %d %d\n",ans,l,r,mid);
if (ans == C)
{
res = mid;
r = mid - 1;
}
else l = mid + 1;
}
printf("%d\n",res);
}
return 0;
}