【费用流】[BZOJ1070]/[HYSBZ1070]修车

题目
分析：将一个技术人员拆成n个点，分别和车连边，第j个技术人员拆成的第k个点修理第i辆车的费用为k*tm[i][j].
当一辆车是这个技术人员修理的倒数第k辆车时，它对等待时间的总数贡的献为k*tm[i][j].
最后，将总数/n即可.

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define MAXN 70
#define MAXM 15
#define INF 0x7f7f7f7f
using namespace std;
queue<int>q;
struct node{
    int u,v,wt,cap;
    node *next,*back;
}*adj[MAXN*MAXM+10],edge[MAXN*MAXM*MAXN*3+10],*ecnt=edge,*pre[MAXN*MAXM+MAXN+10];
int n,m,tm[MAXN+5][MAXM+5],S,T,dist[MAXN*MAXM+MAXN+10],ans;
bool vis[MAXN*MAXM+MAXN+10];
void addedge(int u,int v,int wt,int cap){
    node *p=++ecnt;
    p->v=v;
    p->wt=wt;
    p->cap=cap;
    p->next=adj[u];
    adj[u]=p;
    p=p->back=++ecnt;
    p->back=ecnt-1;
    p->v=u;
    p->wt=-wt;
    p->cap=0;
    p->next=adj[v];
    adj[v]=p;
}
void Read(int &x){
    char c;
    while(c=getchar(),c!=EOF)
        if(c>='0'&&c<='9'){
            x=c-'0';
            while(c=getchar(),c>='0'&&c<='9')
                x=x*10+c-'0';
            ungetc(c,stdin);
            return;
        }
}
void read(){
    Read(m),Read(n);
    int i,j,k;
    T=n*m+n+1;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            Read(tm[i][j]);
    for(k=1;k<=n;k++)
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                addedge((j-1)*n+k,i+n*m,k*tm[i][j],1);
    for(k=1;k<=n;k++)
            for(j=1;j<=m;j++)
                addedge(S,(j-1)*n+k,0,1);
    for(i=1;i<=n;i++)
        addedge(i+n*m,T,0,1);
}
bool spfa(){
    memset(dist,0x7f,sizeof dist);
    memset(pre,0,sizeof pre);
    q.push(S);
    dist[S]=0;
    int u,v;
    vis[S]=1;
    while(!q.empty()){
        u=q.front();
        q.pop();
        vis[u]=0;
        for(node *p=adj[u];p;p=p->next){
            v=p->v;
            if(p->cap>0&&dist[u]+p->wt<dist[v]){
                dist[v]=dist[u]+p->wt;
                pre[v]=p;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    if(dist[T]==INF)
        return 0;
    return 1;
}
void mcmf(){
    int u,delta;
    while(spfa()){
        delta=INF;
        for(u=T;u!=S;u=pre[u]->back->v)
            delta=min(delta,pre[u]->cap);
        for(u=T;u!=S;u=pre[u]->back->v){
            pre[u]->cap-=delta;
            pre[u]->back->cap+=delta;
            ans+=pre[u]->wt*delta;
        }
    }
}
int main()
{
    read();
    mcmf();
    printf("%.2lf\n",1.0*ans/n);
}