分析：巧妙构图，记得某一次bc的最后一道题就是这个，就是借鉴这个老题吧，貌似很多题都是借鉴这个题
关键在于，倒着看是某个人倒数第几个修，这样只影响后面的人，跑费用流就好了
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <vector> using namespace std; const int maxn=1200; const int INF=0x3f3f3f3f; struct Edge { int from,to,cap,flow,cost; Edge(int u,int v,int c,int d,int p):from(u),to(v),cap(c),flow(d),cost(p) {} }; struct MCMF { int s,t,flow,cost; vector<Edge>edges; vector<int>G[maxn]; int inq[maxn]; int d[maxn]; int p[maxn]; int a[maxn]; void init() { for(int i=0;i<maxn;i++) G[i].clear(); edges.clear(); } void AddEdge(int from,int to,int cap,int cost) { edges.push_back((Edge){from,to,cap,0,cost}); edges.push_back((Edge){to,from,0,0,-cost}); int l=edges.size(); G[from].push_back(l-2); G[to].push_back(l-1); } bool BellmanFord(int s,int t) { for(int i=0;i<maxn;i++) d[i]=INF; memset(inq,0,sizeof(inq)); d[s]=0;inq[s]=1;p[s]=0;a[s]=INF; queue<int>Q; Q.push(s); while(!Q.empty()) { int u=Q.front(); Q.pop(); inq[u]=0; for(int i=0;i<G[u].size();i++) { Edge &e=edges[G[u][i]]; if(e.cap>e.flow&&d[e.to]>d[u]+e.cost) { d[e.to]=d[u]+e.cost; p[e.to]=G[u][i]; a[e.to]=min(a[u],e.cap-e.flow); if(!inq[e.to]) { Q.push(e.to); inq[e.to]=1; } } } } if(d[t]==INF)return false; flow+=a[t]; cost+=d[t]*a[t]; int u=t; while(u!=s) { edges[p[u]].flow+=a[t]; edges[p[u]^1].flow-=a[t]; u=edges[p[u]].from; } return true; } int Mincost(int s,int t) { flow=cost=0; while(BellmanFord(s,t)); return cost; } }solve; int c[65][65]; int main(){ int m,n; scanf("%d%d",&m,&n); for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) scanf("%d",&c[i][j]); solve.init(); for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ for(int k=1;k<=n;++k){ int pos=(j-1)*n+k; solve.AddEdge(i+n*m,pos,1,k*c[i][j]); } } } for(int i=1;i<=n;++i)solve.AddEdge(0,i+n*m,1,0); for(int i=1;i<=n*m;++i)solve.AddEdge(i,n+n*m+1,1,0); printf("%.2f ",1.0*solve.Mincost(0,n+n*m+1)/n); return 0; }