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=300;
const int oo=1000000000;

int n,m,k;
int a[maxn],b[maxn];
int ans=0;

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=0;e.cost=w;
    edges.push_back(e);
    e.from=y;e.to=x;e.cap=0;e.flow=0;e.cost=-w;
    edges.push_back(e);
    int c=edges.size();
    G[x].push_back(c-2);
    G[y].push_back(c-1);
}

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=1;i<=totn;++i){
        d[i]=oo;inq[i]=p[i]=0;
    }
    d[s]=0;q.push(s);
    while(!q.empty()){
        int x=q.front();q.pop();inq[x]=0;
        for(int i=0;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]=1;
                }
            }
        }
    }
    
    if(d[t]==oo)return 0;
    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]^1].flow-=f;
        x=edges[p[x]].from;
    }
    return 1;
}

int Mincost(){
    int flow=0,cost=0;
    while(Spfa(flow,cost)){
    }
    return cost;
}

char ss[200];
int main(){
    scanf("%d%d%d",&n,&m,&k);
    totn=n+m;s=++totn;t=++totn;
    for(int i=1;i<=m;++i)scanf("%d",&a[i]);
    for(int i=1;i<=m;++i)scanf("%d",&b[i]);
    for(int i=1;i<=m;++i){
        for(int j=1;j<=n;++j){
            Addedge(i+n,t,1,(2*j-1)*a[i]-b[i]);
        }
    }
    for(int i=1;i<=n;++i){
        scanf("%s",ss+1);
        for(int j=1;j<=m;++j){
            if(ss[j]=='1')Addedge(i,j+n,1,0);
        }
    }
    for(int i=1;i<=n;++i)Addedge(s,i,k,0);
    for(int i=1;i<=n;++i)Addedge(i,t,k-1,0);
    cout<<Mincost()<<endl;
    return 0;
}