★ Title Description
N clients have to wait for the car service, and now the M maintenance staff, maintenance workers repair each different time for each car,
Please arrange each service member in need of repair and car maintenance order, so that all customers always wait a minimum of time.
★ input format
The first two digital input line M, N (1 <= N, M <= 20,1 <= K <= 20), indicates the number of vehicles and the number of maintenance staff.
Next N lines M digital Ti, j (1 <= Ti, j <= 20) represents the j-th vehicle maintenance worker needs repair i-Ti, j time
★ output format
The total output of the minimum waiting time.
★ sample input
2 2
3 2
1 4★ Sample Output
3★ Tips
no
★ reference code
/*
这道题是最小费用网络
现在问题是要如何构建这个网络图
如何表达一辆车对总等待时间的贡献？
第i号车是第j位维修员修理的倒数第k辆车，那么对总时间的贡献 T = k*T[i,j]
（为了表达简洁，这里用“倒数第k辆车”，这样对结果没有影响）
网络图如上所示
现在就是依据这张图，求最大流量的最小费用
*/
#include<bits/stdc++.h>
using namespace std;
int M,N;
int T[20+5][20+5];
struct Edge{
int to,next,flow,dis;//flow流量 dis花费
}E[10000]; //保存的数据是边
int H[10000], numEdge;
void AddEdge(int u, int v, int f, int d){
++numEdge;
E[numEdge].to=v;
E[numEdge].flow=f;
E[numEdge].dis=d;
E[numEdge].next=H[u];
H[u]=numEdge;
}
bool vis[10000];
int dis[10000],pre[10000],last[10000],flow[10000];//dis最小花费;pre每个点的前驱；last每个点的所连的前一条边；flow源点到此处的流量
int maxflow,mincost;
queue <int> q; //保存的数据是点（从起点到终点）
//使用深度优先DFS方法查找所有的增广路径
int spfa(int s,int t){
memset(dis,0x7f,sizeof(dis));
memset(flow,0x7f,sizeof(flow));
memset(vis,0,sizeof(vis));
q.push(s); vis[s]=1; dis[s]=0; pre[t]=-1;
while (!q.empty()) {
int now=q.front(); q.pop();
vis[now]=0;
for(int i=H[now]; i!=-1; i=E[i].next) {//从前端点now出发连着很多的后端点i
if(E[i].flow>0 && dis[E[i].to]>dis[now]+E[i].dis) { //第i条边还有余量，且会使费用更小
dis[E[i].to]=dis[now]+E[i].dis;
pre[E[i].to]=now; //每个点所连的前端点是now
last[E[i].to]=i; //每个点的所连的前一条边
flow[E[i].to]=min(flow[now],E[i].flow);//最大流量取决于最小容量
if(!vis[E[i].to]) {
vis[E[i].to]=1;
q.push(E[i].to);
}
}
}
}
return pre[t];
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cin>>M>>N;
int sta=0, end=N+M*N+1; //“超级源点”和“超级结点”
for(int i=1; i<=N; ++i) for(int j=1; j<=M; ++j){
cin>>T[i][j];
}
//构建网络图
memset(H,-1,sizeof(H)), numEdge=-1;
for(int i=1; i<=N; ++i){
AddEdge(sta, i, 1, 0); //从“超级源点”到“第i辆车”
AddEdge(i, sta, 0, 0);
for(int j=1; j<=M; ++j) for(int k=1; k<=N; ++k){
AddEdge(i, j*N+k, 1, k*T[i][j]); //从“第i辆车”到“第j个维修员修的倒数第k辆车”
AddEdge(j*N+k, i, 0, -k*T[i][j]);
if(i==N){
AddEdge(j*N+k, end, 1, 0); //从“第j个维修员修的倒数第k辆车”到“超级结点”
AddEdge(end, j*N+k, 0, 0);
}
}
}
//求最大流最小费用
int maxflow=0, mincost=0;
while(spfa(sta, end)!=-1) { //迭代，不断的更新网络
maxflow+=flow[end];
mincost+=flow[end]*dis[end];
int now=end;
while(now!=sta) {//从源点一直回溯到汇点
E[last[now]].flow-=flow[end];//flow和dis容易搞混
E[last[now]^1].flow+=flow[end];
now=pre[now];
}
}
cout<<mincost<<endl;
return 0;
}