Optimal Milking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 10010 | Accepted: 3630 | |
| Case Time Limit: 1000MS | ||
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
Source
-----------求最大值最小，二分最大值建图判断是否成立即可。
----------
/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) {
REP(i, n){
REP(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
#define Display_1(A, n, m) {
REP_1(i, n){
REP_1(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxm=11111;
const int maxn=333;
struct edgenode
{
int to,flow,next;
};
struct Dinic
{
int node,src,dest,edge;
int head[maxn],work[maxn],dis[maxn],q[maxn];
edgenode edges[maxm];
void prepare(int _node,int _src,int _dest)
{
node=_node,src=_src,dest=_dest;
for (int i=0; i<node; i++) head[i]=-1;
edge=0;
}
void addedge(int u,int v,int c)
{
edges[edge].flow=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
edges[edge].flow=0,edges[edge].to=u,edges[edge].next=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for (i=0; i<node; i++) dis[i]=-1;
dis[q[r++]=src]=0;
for (l=0; l<r; l++){
for (i=head[u=q[l]]; i!=-1; i=edges[i].next){
if (edges[i].flow&&dis[v=edges[i].to]<0){
dis[q[r++]=v]=dis[u]+1;
if (v==dest) return true;
}
}
}
return false;
}
int Dinic_dfs(int u,int exp)
{
if (u==dest) return exp;
for (int &i=work[u],v,tmp; i!=-1; i=edges[i].next){
if (edges[i].flow&&dis[v=edges[i].to]==dis[u]+1&&
(tmp=Dinic_dfs(v,min(exp,edges[i].flow)))>0){
edges[i].flow-=tmp;
edges[i^1].flow+=tmp;
return tmp;
}
}
return 0;
}
int Dinic_flow()
{
int i,ret=0,delta;
while (Dinic_bfs()){
for (i=0; i<node; i++) work[i]=head[i];
while ( delta=Dinic_dfs(src,INF) ) ret+=delta;
}
return ret;
}
}solver;
int a[maxn][maxn];
int f[maxn][maxn];
int n,m;
int K,C,M;
bool CS(int mid)
{
solver.prepare(n+2,0,n+1);
FOR_1(i,1,K) solver.addedge(0,i,M);
FOR_1(i,K+1,K+C) solver.addedge(i,n+1,1);
FOR_1(i,1,K)FOR_1(j,K+1,K+C)
if (f[i][j]!=0&&f[i][j]<=mid)
solver.addedge(i,j,1);
int ret=solver.Dinic_flow();
if (ret==C) return true;
else return false;
}
int main()
{
int ans;
while (~scanf("%d%d%d",&K,&C,&M))
{
n=K+C;
REP_1(i,n) REP_1(j,n)
{
scanf("%d",&a[i][j]);
f[i][j]=a[i][j];
if (f[i][j]==0) f[i][j]=INF;
}
REP_1(k,n)REP_1(i,n)REP_1(j,n)
if (f[i][k]!=INF&&f[k][j]!=INF&&f[i][k]+f[k][j]<f[i][j])
f[i][j]=f[i][k]+f[k][j];
int l,r;
l=0;
r=200*(K+C);
ans=0;
while (l<=r)
{
int mid=l+(r-l)/2;
if (CS(mid))
{
ans=mid;
r=mid-1;
}
else
{
l=mid+1;
}
}
printf("%d
",ans);
}
return 0;
}