Poj Optimal Milking 2112 floyd+dinic+二分
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.
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.
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
分析：
本题要安排C头奶牛到某个挤奶器，使得每头奶牛需要走的路程中最大路程的距离最小。
本题的求解方法:先用floyd算法求出能达到的任意两点之间的最短路径，然后用dinic算法求最大流；搜索最大距离的最小值采用二分法进行。
代码如下：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAX = 300;
int dis[MAX][MAX];
int map[MAX][MAX];
bool sign[MAX][MAX];
bool used[MAX];
int K,C,n,M;
queue<int > q;
void floyd() {
for(int k = 1;k <= n;k++) {
for(int i = 1;i <= n;i++) {
if(dis[i][k] == INF) continue;
for(int j = 1;j <= n;j++) {
if(dis[k][j] == INF) continue;
dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
}
}
}
}
void Build_Graph(int min_max) {
memset(map,0,sizeof(map));
for(int i = 1;i <= n;i++) {
if(i<=K) map[i][n+1] = M;
else map[0][i] = 1;
}
for(int i = K+1;i<=n;i++) {
for(int j = 1;j <=K;j++) {
if(dis[i][j] <= min_max) map[i][j] = 1;
}
}
}
bool bfs() {
memset(used,0,sizeof(used));
memset(sign,0,sizeof(sign));
while(!q.empty()) q.pop();
used[0] = 1;
q.push(0);
while(!q.empty()) {
for(int i = 0;i <= n+1;i++) {
if(!used[i] && map[q.front()][i]) {
q.push(i);
used[i] = 1;
sign[q.front()][i] = 1;
}
}
q.pop();
}
if(used[n+1]) return true;
else return false;
}
int dfs(int v,int sum) {
int s,t;
if(v == n+1) return sum;
s = sum;
for(int i = 0;i <= n+1;i++) {
if(sign[v][i]) {
t = dfs(i,min(map[v][i],sum));
map[v][i] -= t;
map[i][v] += t;
sum -= t;
}
}
return s - sum;
}
int dinic() {
int sum = 0;
while(bfs()) sum+=dfs(0,INF);
return sum;
}
int main() {
int l,r,mid,ans;
scanf("%d%d%d",&K,&C,&M);
n = K+C;
memset(dis,0x3f,sizeof(dis));
for(int i = 1;i <= n;i++) {
for(int j = 1;j <= n;j++) {
scanf("%d",&dis[i][j]);
if(i !=j && dis[i][j] == 0) dis[i][j] = INF;
}
}
floyd();
l = 0,r = 10000;
while(l<r) {
mid = (l+r)/2;
Build_Graph(mid);
ans = dinic();
if(ans>=C) r = mid;
else l = mid+1;
}
printf("%d\n",r);
return 0;
}转载于:https://www.cnblogs.com/pot-a-to/p/10957291.html