No.22 - POJ2112 网络流求二部图分配方案

首先floyd求最短路
然后构建网络流，其中源点到每一头牛容量为1，每一个挤奶器到汇点容量为m
牛到挤奶器容量为1.

二分答案，构建网络流。

// ShellDawn
// POJ2112
// No.22

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
#define MM(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
using namespace std;

#define maxn 300
int n;
int D[maxn][maxn];
int E[maxn][maxn];
int V[maxn];
int P[maxn][maxn][2];
int C[maxn];

// 1~c牛，c+1～k+c挤奶器
void init(int gate,int k,int c,int m){
    MM(E,0);
    for(int i=1;i<=c;i++) E[0][i] = 1;
    for(int i=c+1;i<=c+k;i++) E[i][n+1] = m;
    for(int i=1;i<=c;i++){
        for(int j=c+1;j<=c+k;j++){
            if(D[i][j] <= gate){
                E[i][j] = 1;
            }
        }
    }
}

bool BFS(int s){
    MM(V,0);MM(C,0);
    queue<int> q;
    q.push(s);
    V[s] = 1;
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i=1;i<=n+1;i++){
            if(V[i] == 0 && E[now][i] > 0){
                q.push(i);
                V[i] = V[now]+1;
                int c = C[now];
                P[now][c][0] = i;
                P[now][c][1] = E[now][i];
                C[now]++;
                if(i == n+1) return true;
            }
        }
    }
    return false;
}

int DFS(int now,int minflow){
    if(now == n+1) return minflow;
    int flow = 0;
    for(int i=0;i<C[now];i++){
        int next = P[now][i][0];
        int v = P[now][i][1];
        int f = DFS(next,min(minflow,v));
        flow += f;
        minflow -= f;
        E[now][next] -= f;
        E[next][now] += f;
    }
    return flow;
}

int main(){
    //freopen("A.txt","r",stdin);
    int k,c,m;
    cin>>k>>c>>m;
    n = k + c;
    MM(D,0x3f);
    for(int i=c+1;i<=k+c;i++){
        for(int j=c+1;j<=k+c;j++){
            scanf("%d",&D[i][j]);
        }
        for(int j=1;j<=c;j++){
            scanf("%d",&D[i][j]);
        }
    }
    for(int i=1;i<=c;i++){
        for(int j=c+1;j<=k+c;j++){
            scanf("%d",&D[i][j]);
        }
        for(int j=1;j<=c;j++){
            scanf("%d",&D[i][j]);
        }
    }
    // floyd
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(D[i][j] == 0) D[i][j] = INF;
        }
    }
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                D[i][j] = min(D[i][j],D[i][k] + D[k][j]);
            }
        }
    }
    // binary search
    int L = 0;
    int R = 10001;
    int ans = INF;
    while(L<R){
        int mid = (L+R)/2;
        init(mid,k,c,m);
        int flow = 0;
        while(BFS(0)) flow+=DFS(0,INF);
        if(flow == c){
            ans = min(ans,mid);
            R = mid;
        }else L = mid + 1;
    }
    cout<<ans<<endl;
    return 0;
}