Optimal Milking
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.
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.
* 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
题意：有k个牛奶机跟c头牛。他们之间有路相连，农民想让每个牛能到其中一个牛奶机，又想让最累的牛累的程度最小。
分析：本来傻傻的用费用流，但发现不对，因为每个增广路不是单纯的一个牛走的路，会包含其他牛走过的路在里面，详细另一篇关于最大流的归纳。
这个题是求牛走的最长路，那枚举ans，然后对牛到奶牛机的最短距离满足ans的连一条容量inf的边，如果最大流等于牛的数量，则ans是合法的。
二分最大流代码：
#include < stdio.h >
#include < string .h >
#include < algorithm >
#define Min(a, b) (a) < (b) ? a : b
using namespace std;
const int MAXN = 1005 ;
const int MAXM = 210000 ;
const int INF = 1000000000 ;
struct Edge
{
int st, ed;
int next;
int flow;
int cap;
}edge[MAXM];
int head[MAXN], level[MAXN], E, map[MAXN][MAXN];
void add( int u, int v, int w)
{
// printf("add %d %d %d\n", u, v, w);
edge[E].flow = 0 ;
edge[E].cap = w;
edge[E].st = u;
edge[E].ed = v;
edge[E].next = head[u];
head[u] = E ++ ;
edge[E].flow = 0 ;
edge[E].cap = 0 ;
edge[E].st = v;
edge[E].ed = u;
edge[E].next = head[v];
head[v] = E ++ ;
}
int dinic_bfs( int src, int dest, int ver)
{
int i, j;
for (i = 0 ; i <= ver; i ++ )
{
level[i] = - 1 ;
}
int que[MAXN], rear = 1 ;
que[ 0 ] = src; level[src] = 0 ;
for (i = 0 ; i < rear; i ++ )
{
for (j = head[que[i]]; j != - 1 ; j = edge[j].next)
{
if (level[edge[j].ed] == - 1 && edge[j].cap > edge[j].flow)
{
level[edge[j].ed] = level[que[i]] + 1 ;
que[rear ++ ] = edge[j].ed;
}
}
}
return level[dest] >= 0 ;
}
int dinic_dfs( int src, int dest, int ver)
{
int stk[MAXN], top = 0 ;
int ret = 0 , cur, ptr, pre[MAXN], minf, i;
int del[MAXN];
for (i = 0 ; i <= ver; i ++ )
{
del[i] = 0 ;
}
stk[top ++ ] = src;
pre[src] = src;
cur = src;
while (top)
{
while (cur != dest && top)
{
for (i = head[cur]; i != - 1 ; i = edge[i].next)
{
if (level[edge[i].ed] == level[cur] + 1 && edge[i].cap > edge[i].flow && ! del[edge[i].ed])
{
stk[top ++ ] = edge[i].ed;
cur = edge[i].ed;
pre[edge[i].ed] = i;
break ;
}
}
if (i == - 1 )
{
del[cur] = 1 ;
top -- ;
if (top) cur = stk[top - 1 ];
}
}
if (cur == dest)
{
minf = INF;
while (cur != src)
{
cur = pre[cur];
if (edge[cur].cap - edge[cur].flow < minf) minf = edge[cur].cap - edge[cur].flow;
cur = edge[cur].st;
}
cur = dest;
while (cur != src)
{
cur = pre[cur];
edge[cur].flow += minf;
edge[cur ^ 1 ].flow -= minf;
if (edge[cur].cap - edge[cur].flow == 0 )
{
ptr = edge[cur].st;
}
cur = edge[cur].st;
}
while (top > 0 && stk[top - 1 ] != ptr) top -- ;
if (top) cur = stk[top - 1 ];
ret += minf;
}
}
return ret;
}
int Dinic( int src, int dest, int ver)
{
int ret = 0 , t;
while (dinic_bfs(src, dest, ver))
{
t = dinic_dfs(src, dest, ver);
if (t) ret += t;
else break ;
}
return ret;
}
void floyd( int n) // 每个点的最近距离,这个不多说了。保证map[i][k]跟map[k][j]都以存在就行
{
int k, i, j;
for (k = 1 ; k <= n; k ++ ) // k表示中间点，一定放在最外面
{
for (i = 1 ; i <= n; i ++ )
{
for (j = 1 ; j <= n; j ++ )
{
if (i != j && map[i][k] != INF && map[k][j] != INF && map[i][k] + map[k][j] < map[i][j])
{
map[i][j] = map[i][k] + map[k][j];
}
}
}
}
}
void build( int distance, int k, int c, int m)
{
int n = c + k, i, j;
E = 0 ;
for (i = 0 ; i <= n + 1 ; i ++ )
{
head[i] = - 1 ;
}
for (i = k + 1 ; i <= n; i ++ )
{
add( 0 , i, 1 ); // 源到牛的容量是1 ，题目是要求每只牛都能去任意一个奶牛机就行了
for (j = 1 ; j <= k; j ++ )
{
if (map[i][j] <= distance) // 对满足距离的两个点牛跟牛奶机
{
add(i, j, INF); // 牛到牛奶机的容量inf
}
}
}
for (j = 1 ; j <= k; j ++ ) // 牛奶机到汇的容量为m
{
add(j, n + 1 , m);
}
}
int main()
{
int c, k, m, n, i, j;
scanf( " %d%d%d " , & k, & c, & m);
n = c + k;
for (i = 1 ; i <= n; i ++ )
{
for (j = 1 ; j <= n; j ++ )
{
scanf( " %d " , & map[i][j]);
if (map[i][j] == 0 )
{
map[i][j] = INF;
}
}
}
floyd(n);
int l = 0 , r = INF - 10 , mid, ans;
int s = 0 , t = c + k + 1 , ver = t + 1 ;
while (l < r) // 因为二分出现l+1=r的时候，(l+r) >> 1 = l的。所以l+1就等于r了，当l跟r相等的时候就是解
{
mid = (l + r) >> 1 ;
build(mid, k, c, m);
ans = Dinic( 0 , t, ver);
if (ans == c)
{
r = mid; // r是保证能满足最大流等于牛的数量
}
else
{
l = mid + 1 ;
}
}
printf( " %d\n " , r);
return 0 ;
}
/*
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
1 1 1
0 1
1 0
2 2 1
0 0 1 3
0 0 2 100
1 2 0 0
3 100 0 0
*/