BZOJ 3442 学习小组

题解：

神建图

普通的二分图费用流建完后

添加学生x->t 容量为k-1的边

表示尽量让x参加一个活动，剩下的k-1次机会可以不参加

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn=;
const int oo=;

int n,m,k;
int a[maxn],b[maxn];
int ans=;

struct Edge{
    int from,to,cap,flow,cost;
};
vector<int>G[maxn];
vector<Edge>edges;
void Addedge(int x,int y,int z,int w){
    Edge e;
    e.from=x;e.to=y;e.cap=z;e.flow=;e.cost=w;
    edges.push_back(e);
    e.from=y;e.to=x;e.cap=;e.flow=;e.cost=-w;
    edges.push_back(e);
    int c=edges.size();
    G[x].push_back(c-);
    G[y].push_back(c-);
}

int s,t,totn;
queue<int>q;
int inq[maxn];
int d[maxn];
int p[maxn];

int Spfa(int &nowflow,int &nowcost){
    for(int i=;i<=totn;++i){
        d[i]=oo;inq[i]=p[i]=;
    }
    d[s]=;q.push(s);
    while(!q.empty()){
        int x=q.front();q.pop();inq[x]=;
        for(int i=;i<G[x].size();++i){
            Edge e=edges[G[x][i]];
            if((e.cap>e.flow)&&(d[x]+e.cost<d[e.to])){
                d[e.to]=d[x]+e.cost;
                p[e.to]=G[x][i];
                if(!inq[e.to]){
                    q.push(e.to);
                    inq[e.to]=;
                }
            }
        }
    }

    if(d[t]==oo)return ;
    int x=t,f=oo;
    while(x!=s){
        Edge e=edges[p[x]];
        f=min(f,e.cap-e.flow);
        x=e.from;
    }
    nowflow+=f;nowcost+=f*d[t];
    x=t;
    while(x!=s){
        edges[p[x]].flow+=f;
        edges[p[x]^].flow-=f;
        x=edges[p[x]].from;
    }
    return ;
}

int Mincost(){
    int flow=,cost=;
    while(Spfa(flow,cost)){
    }
    return cost;
}

char ss[];
int main(){
    scanf("%d%d%d",&n,&m,&k);
    totn=n+m;s=++totn;t=++totn;
    for(int i=;i<=m;++i)scanf("%d",&a[i]);
    for(int i=;i<=m;++i)scanf("%d",&b[i]);
    for(int i=;i<=m;++i){
        for(int j=;j<=n;++j){
            Addedge(i+n,t,,(*j-)*a[i]-b[i]);
        }
    }
    for(int i=;i<=n;++i){
        scanf("%s",ss+);
        for(int j=;j<=m;++j){
            if(ss[j]=='')Addedge(i,j+n,,);
        }
    }
    for(int i=;i<=n;++i)Addedge(s,i,k,);
    for(int i=;i<=n;++i)Addedge(i,t,k-,);
    cout<<Mincost()<<endl;
    return ;
}