BZOJ 1070 [SCOI2007]修车

题解：

最小费用最大流

为每个师傅建一排点，表示这个师傅修的第几量车，然后他会对后面的人产生影响

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=10009;
const int oo=1000000000;

int n,m;
int p[200][200];

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;
int pre[maxn];
int d[maxn];
int inq[maxn];
queue<int>q;
int Spfa(int &nowflow,int &nowcost){
    for(int i=1;i<=totn;++i){
        inq[i]=0;d[i]=oo;
    }
    inq[s]=1;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;
                pre[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[pre[x]];
        f=min(f,e.cap-e.flow);
        x=e.from;
    }
    nowflow+=f;nowcost+=d[t]*f;
    x=t;
    while(x!=s){
        edges[pre[x]].flow+=f;
        edges[pre[x]^1].flow-=f;
        x=edges[pre[x]].from;
    }
    return 1;
}

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

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        for(int j=1;j<=m;++j){
            p[i][j]=++totn;
        }
    }
    s=++totn;t=++totn;
    for(int j=1;j<=m;++j){
        for(int i=1;i<=n;++i){
            int x;scanf("%d",&x);
            for(int k=1;k<=m;++k){
                Addedge(p[i][k],totn+j,1,k*x);
            }
        }
    }

    for(int i=1;i<=n*m;++i)Addedge(s,i,1,0);
    for(int i=totn+1;i<=totn+m;++i)Addedge(i,t,1,0);
    totn+=m;
    
    printf("%.2f\n",Mincost()*1.0/m);
    return 0;
}