POJ 2112 Optimal Milking
题意：农场主John 将他的K(1≤K≤30)个挤奶器运到牧场，在那里有C(1≤C≤200)头奶牛，在奶牛和挤奶器之间有一组不同长度的路。K个挤奶器的位置用1～K的编号标明，奶牛的位置用K+1～K+C 的编号标明。每台挤奶器每天最多能为M(1≤M≤15)头奶牛挤奶。寻找一个方案，安排每头奶牛到某个挤奶器挤奶，并使得C 头奶牛须要走的全部路程中的最大路程最小。每一个測试数据中至少有一个安排方案。每条奶牛到挤奶器有多条路。
思路：先用Floyd 算法求出能达到的随意两点之间的最短路径，然后二分最大距离的最小值，每次用二分的值求最大流。
在求最大流时构图：建立一个源点，每一个点到挤奶器连一条流量为m的边。建立一个汇点，每头奶牛到汇点连一条流量为1的边。挤奶器与奶牛之间的距离小于等于mid则连边，流量为1。最后求最大流是否为c就可以。
代码：
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
//#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 250;
const int maxm = 100000;
int k, c, m;
int mp[maxn][maxn];
struct node {
int v; // vertex
int cap; // capacity
int flow; // current flow in this arc
int nxt;
} e[maxm * 2];
int g[maxn], cnt;
int st, ed, n;
void add(int u, int v, int c) {
e[++cnt].v = v;
e[cnt].cap = c;
e[cnt].flow = 0;
e[cnt].nxt = g[u];
g[u] = cnt;
e[++cnt].v = u;
e[cnt].cap = 0;
e[cnt].flow = 0;
e[cnt].nxt = g[v];
g[v] = cnt;
}
void init(int mid) {
mem(g, 0);
cnt = 1;
st = 0, ed = k + c + 1;
n = k + c;
for (int i = 1; i <= k; i++) add(st, i, m);
for (int i = k + 1; i <= n; i++) {
for (int j = 1; j <= k; j++) if (mp[i][j] <= mid) add(j, i, 1);
add(i, ed, 1);
}
n += 3;
}
int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
int font = 0, rear = 1;
for (int i = 0; i <= n; i++) { //n为总点数
dist[i] = maxn;
numbs[i] = 0;
}
q[font] = ed;
dist[ed] = 0;
numbs[0] = 1;
while(font != rear) {
int u = q[font++];
for (int i = g[u]; i; i = e[i].nxt) {
if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
dist[e[i].v] = dist[u] + 1;
++numbs[dist[e[i].v]];
q[rear++] = e[i].v;
}
}
}
int maxflow() {
rev_bfs();
int u, totalflow = 0;
int curg[maxn], revpath[maxn];
for(int i = 0; i <= n; ++i) curg[i] = g[i];
u = st;
while(dist[st] < n) {
if(u == ed) { // find an augmenting path
int augflow = INF;
for(int i = st; i != ed; i = e[curg[i]].v)
augflow = min(augflow, e[curg[i]].cap);
for(int i = st; i != ed; i = e[curg[i]].v) {
e[curg[i]].cap -= augflow;
e[curg[i] ^ 1].cap += augflow;
e[curg[i]].flow += augflow;
e[curg[i] ^ 1].flow -= augflow;
}
totalflow += augflow;
u = st;
}
int i;
for(i = curg[u]; i; i = e[i].nxt)
if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
if(i) { // find an admissible arc, then Advance
curg[u] = i;
revpath[e[i].v] = i ^ 1;
u = e[i].v;
} else { // no admissible arc, then relabel this vertex
if(0 == (--numbs[dist[u]])) break; // GAP cut, Important!
curg[u] = g[u];
int mindist = n;
for(int j = g[u]; j; j = e[j].nxt)
if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
dist[u] = mindist + 1;
++numbs[dist[u]];
if(u != st)
u = e[revpath[u]].v; // Backtrack
}
}
return totalflow;
}
void get() {
n = k + c;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
scanf("%d", mp[i] + j);
if (mp[i][j] == 0 && i != j) mp[i][j] = INF;
}
}
}
void floyd(int n, int mp[][maxn]) {
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++) if (mp[i][k] != INF) {
for (int j = 1; j <= n; j++) mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
}
}
}
int main () {
while(scanf("%d%d%d", &k, &c, &m) != EOF) {
get();
floyd(n, mp);
int l = 0, r = 10000, mid;
while(l < r) {
mid = (l + r) >> 1;
init(mid);
//debug;
if (maxflow() >= c) r = mid;
else l = mid + 1;
}
printf("%d
", r);
}
return 0;
}