Optimal Milking

Time Limit: 2000MS	Memory Limit: 30000K
Total Submissions: 6483	Accepted: 2428
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. 

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

Source

分析：這題一開始不知從何下手，後來發現只要把所有點之間的最短路徑都做成一條容量為1邊就行，然後增加源和匯，源與所有機器連上容量為m的邊，牛與匯連上容量為1的邊，然後二分答案，大於答案的邊都刪掉，判斷是否可行即可。。。 程式碼：

#include<cstdio>
using namespace std;
const int mm=55555;
const int mn=333;
const int oo=1000000000;
int node,src,dest,edge;
int ver[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn];
int d[mn][mn];
inline int min(int a,int b)
{
    return a<b?a:b;
}
inline 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;
}
inline void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,next[edge]=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>=0;i=next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp;i>=0;i=next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=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,oo))ret+=delta;
    }
    return ret;
}
int main()
{
    int u,v,w,n,c,m,s,l,r,mid;
    while(scanf("%d%d%d",&n,&c,&m)!=-1)
    {
        s=n+c;
        for(u=1;u<=s;++u)
            for(v=1;v<=s;++v)
                scanf("%d",&d[u][v]);
        for(w=1;w<=s;++w)
            for(u=1;u<=s;++u)
                if(w!=u)for(v=1;v<=s;++v)
                    if(u!=v&&w!=v&&d[u][w]&&d[w][v])
                    {
                        if(d[u][v])d[u][v]=min(d[u][v],d[u][w]+d[w][v]);
                        else d[u][v]=d[u][w]+d[w][v];
                    }
        l=oo,r=0;
        for(u=1;u<=s;++u)
            for(v=1;v<=s;++v)
            if(d[u][v])
            {
                if(d[u][v]<l)l=d[u][v];
                if(d[u][v]>r)r=d[u][v];
            }
        while(l<r)
        {
            mid=(l+r)>>1;
            prepare(s+2,0,s+1);
            for(u=1;u<=n;++u)addedge(src,u,m);
            for(u=n+1;u<=s;++u)addedge(u,dest,1);
            for(u=1;u<=n;++u)
                for(v=n+1;v<=s;++v)
                    if(d[u][v]&&d[u][v]<=mid)addedge(u,v,1);
            if(Dinic_flow()==c)r=mid;
            else l=mid+1;
        }
        printf("%d\n",r);
    }
    return 0;
}