POJ 2112 Optimal Milking (Floyd+二分+最大流)

【题意】有K台挤奶机，C头奶牛，在奶牛和机器间有一组长度不同的路，每台机器每天最多能为M头奶牛挤奶。现在要寻找一个方案，安排每头奶牛到某台机器挤奶，使得C头奶牛中走过的路径长度的和的最大值最小。

挺好的一道题~也算是经典的模型，里面所用到的方法也很经典、常用。

【思路】易知模型其实是个二分图，只是每头奶牛可以去多台机器，所以二分图多重匹配就可以解，因为在练最大流，就拿最大流解了。

先Floyd预处理两两之间的最短距离，然后二分距离，每次以二分的mid为容量限制建图，即保留图中长度<=mid的边，然后添加源汇，源到每头牛连容量为1的边，每台机器到汇点连容量为M的边，然后跑一次最大流，如果满流，说明C头奶牛都可以在mid的情况下被分配，则不断二分找mid的最小值即可。

严格上讲这类题需要拆点的，详见此题。本题不用拆点是因为其本身就是二分图，不存在间接流量边。

[cpp]
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 255;
const int MAXE = 100005;
const int oo = 0x3fffffff;
struct node{
int u, v, flow;
int opp;
int next;
};
struct Dinic{
node arc[MAXE];
int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数
int cur[MAXV]; //当前弧
int q[MAXV]; //bfs建层次图时的队列
int path[MAXE], top; //存dfs当前最短路径的栈
int dep[MAXV]; //各节点层次
void init(int n){
vn = n;
en = 0;
mem(head, -1);
}
void insert_flow(int u, int v, int flow){
arc[en].u = u;
arc[en].v = v;
arc[en].flow = flow;
arc[en].opp = en + 1;
arc[en].next = head[u];
head[u] = en ++;

arc[en].u = v;
arc[en].v = u;
arc[en].flow = 0; //反向弧
arc[en].opp = en - 1;
arc[en].next = head[v];
head[v] = en ++;
}
bool bfs(int s, int t){
mem(dep, -1);
int lq = 0, rq = 1;
dep[s] = 0;
q[lq] = s;
while(lq < rq){
int u = q[lq ++];
if (u == t){
return true;
}
for (int i = head[u]; i != -1; i = arc[i].next){
int v = arc[i].v;
if (dep[v] == -1 && arc[i].flow > 0){
dep[v] = dep[u] + 1;
q[rq ++] = v;
}
}
}
return false;
}
int solve(int s, int t){
int maxflow = 0;
while(bfs(s, t)){
int i, j;
for (i = 1; i <= vn; i ++) cur[i] = head[i];
for (i = s, top = 0;;){
if (i == t){
int mink;
int minflow = 0x3fffffff;
for (int k = 0; k < top; k ++)
if (minflow > arc[path[k]].flow){
minflow = arc[path[k]].flow;
mink = k;
}
for (int k = 0; k < top; k ++)
arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
maxflow += minflow;
top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
i = arc[path[top]].u;
}
for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
int v = arc[j].v;
if (arc[j].flow && dep[v] == dep[i] + 1)
break;
}
if (j != -1){
path[top ++] = j;
i = arc[j].v;
}
else{
if (top == 0) break;
dep[i] = -1;
i = arc[path[-- top]].u;
}
}
}
return maxflow;
}
}dinic;
int map[250][250];
void floyd(int n){
for (int k = 0; k < n; k ++){
for (int i = 0; i < n; i ++){
if(map[i][k] == oo) continue;
for (int j = 0; j < n; j ++){
if (map[j][k] == oo) continue;
if (map[i][j] > map[i][k] + map[k][j])
map[i][j] = map[i][k] + map[k][j];
}
}
}
return ;
}
int go(int mid, int k, int c, int m){
dinic.init(k+c+2);
for (int i = 1; i <= k; i ++){
dinic.insert_flow(k+c+1, i, m);
}
for (int i = k+1; i <= k+c; i ++){
dinic.insert_flow(i, k+c+2, 1);
}
for (int i = 0; i < k; i ++){
for (int j = k; j < k+c; j ++){
if (map[i][j] <= mid){
dinic.insert_flow(i+1, j+1, 1);
}
}
}
return dinic.solve(k+c+1, k+c+2);
}
int BS(int k, int c, int m){
int l = 0, r = 10000000;
while(l < r){
int mid = MID(l,r);
if (go(mid, k, c, m) == c){
r = mid;
}
else{
l = mid + 1;
}
}
return r;
}
int main(){
int k, c, m;
scanf("%d %d %d", &k, &c, &m);
for (int i = 0; i < k+c; i ++){
for (int j = 0; j < k+c; j ++){
scanf("%d", &map[i][j]);
if (map[i][j] == 0)
map[i][j] = oo;
}
}
floyd(k+c);
printf("%d\n", BS(k, c, m));
return 0;
}[/cpp]