POJ2112 Optimal Milking

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 17811		Accepted: 6368
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

USACO 2003 U S Open

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题目给出n头牛和m台机器的两两距离，0表示走不通，和每台机器最多容纳的牛数量，问每头牛都去一台机器，最远的牛到机器的最小距离

思路：先floyd跑出两两之间最短距离，在二分最远距离+二分图多重匹配||最大流验证

二分图多重匹配：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int MAXN=1005;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];
int linker[MAXN][MAXN];
bool used[MAXN];
int linknum[MAXN];
int cap[MAXN];
int mp[MAXN][MAXN];
int N;

bool dfs(int u,int mid)
{
    int v;
    for(v=1; v<=vN; v++)
        if(mp[u][v]<=mid&&!used[v])
        {
            used[v]=true;
            if(linknum[v]<cap[v])
            {
                linker[v][++linknum[v]]=u;
                return true;
            }
            for(int i=1; i<=cap[v]; i++)
                if(dfs(linker[v][i],mid))
                {
                    linker[v][i]=u;
                    return true;
                }
        }
    return false;
}

int hungary(int mid)
{
    int res=0;
    int u;
    memset(linknum,0,sizeof linknum);
    memset(linker,-1,sizeof linker);
    for(u=vN+1; u<=N; u++)
    {
        memset(used,0,sizeof used);
        if(dfs(u,mid))  res++;
    }
    return res;
}

void floyd(){
    for(int k = 1; k <= N; ++k){
        for(int i = 1; i <= N; ++i){
            for(int j = 1; j <= N; ++j){
                mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
            }
        }
    }
}
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&vN,&uN,&k))
    {
        N=uN+vN;
        int mx=-1;
        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)
            {
                scanf("%d",&mp[i][j]);
                mx=max(mx,mp[i][j]);
                if(mp[i][j]==0)
                    mp[i][j]=INF;
            }

      floyd();
        for(int i=1; i<=vN; i++)
            cap[i]=k;
        int l=0,r=INF;
        int ans=0;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(hungary(mid)==uN) ans=mid,r=mid-1;
            else l=mid+1;
        }
        printf("%d\n",ans);
    }
    return 0;
}

最大流

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 500

struct node
{
    int u, v, next, cap;
} edge[MAXN*MAXN];
int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];
int cnt;
int n,m,k;
int mp[MAXN][MAXN];
int N;
void init()
{
    cnt = 0;
    memset(s, -1, sizeof(s));
}

void add(int u, int v, int c)
{
    edge[cnt].u = u;
    edge[cnt].v = v;
    edge[cnt].cap = c;
    edge[cnt].next = s[u];
    s[u] = cnt++;
    edge[cnt].u = v;
    edge[cnt].v = u;
    edge[cnt].cap = 0;
    edge[cnt].next = s[v];
    s[v] = cnt++;
}

bool BFS(int ss, int ee)
{
    memset(d, 0, sizeof d);
    d[ss] = 1;
    queue<int>q;
    q.push(ss);
    while (!q.empty())
    {
        int pre = q.front();
        q.pop();
        for (int i = s[pre]; ~i; i = edge[i].next)
        {
            int v = edge[i].v;
            if (edge[i].cap > 0 && !d[v])
            {
                d[v] = d[pre] + 1;
                q.push(v);
            }
        }
    }
    return d[ee];
}

int DFS(int x, int exp, int ee)
{
    if (x == ee||!exp) return exp;
    int temp,flow=0;
    for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
    {
        int v = edge[i].v;
        if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
        {
            edge[i].cap -= temp;
            edge[i ^ 1].cap += temp;
            flow += temp;
            exp -= temp;
            if (!exp) break;
        }
    }
    if (!flow) d[x] = 0;
    return flow;
}

int Dinic_flow(int mid)
{
    init();
    for(int i=1; i<=n; i++)
        add(0,i,k);
    for(int i=n+1; i<=N; i++)
        add(i,n+1+m,1);
    for(int i=1; i<=n; i++)
        for(int j=1+n; j<=N; j++)
            if(mp[i][j]<=mid)
                add(i,j,1);
    int ss=0,ee=m+n+1;
    int ans = 0;
    while (BFS(ss, ee))
    {
        for (int i = 0; i <=ee; i++) nt[i] = s[i];
        ans+= DFS(ss, INF, ee);
    }
    return ans;
}

void floyd()
{
    for(int k = 1; k <= N; ++k)
    {
        for(int i = 1; i <= N; ++i)
        {
            for(int j = 1; j <= N; ++j)
            {
                mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
            }
        }
    }
}

int main()
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        N=m+n;
        int mx=-1;
        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)
            {
                scanf("%d",&mp[i][j]);
                mx=max(mx,mp[i][j]);
                if(mp[i][j]==0)
                    mp[i][j]=INF;
            }
        floyd();

        int l=0,r=INF;
        int ans=0;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(Dinic_flow(mid)==m) ans=mid,r=mid-1;
            else l=mid+1;
        }
        printf("%d\n",ans);
    }
    return 0;
}