题目链接
题解
假设一个大叔修了k辆车,修的第i个车的车主等待时间是自己车被修的时间的与前边\(i-1\)辆车修车时间的和
它对答案的贡献就是\(x_i*(k-i+1)\)
把大叔拆成n个点，每一层点代表该时刻，顾客连向大叔们，流连为，费用为\((k-层数+1)*等待时间\)
ps ~~n,m看反狂 WA 4次ORZ
代码
// luogu-judger-enable-o2
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using std::min;
using std::queue;
const int maxn = 10;
const int maxm = 35007;
inline int read() {
int x=0;
char c=getchar();
while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
return x;
}
int n,m;
int peo[70][10];
struct node{
int u,v,next,flow,cost;
}edge[maxm<<1];
int num=1,head[maxm];
void add_edge(int u,int v,int flow,int cost) {
edge[++num].v=v;edge[num].u=u;edge[num].next=head[u];edge[num].flow=flow;edge[num].cost=cost;head[u]=num;
edge[++num].v=u;edge[num].u=v;edge[num].next=head[v];edge[num].flow=0;edge[num].cost=-cost;head[v]=num;
}
int dis[maxm<<1],pre[maxm<<1];bool vis[maxm<<1];
int S=0,T;
bool spfa() {
memset(dis,0x3f,sizeof dis);
vis[S]=1;dis[S]=0;
queue<int>que;
que.push(S);
while(!que.empty()) {
int u=que.front();
que.pop();
for(int i=head[u];i;i=edge[i].next) {
int v=edge[i].v;
if(edge[i].flow&&dis[u]+edge[i].cost<dis[v]) {
dis[v]=edge[i].cost+dis[u];
pre[v]=i;
if(!vis[v])
que.push(v),vis[v]=1;
}
}
vis[u]=0;
}
if(dis[T]!=0x3f3f3f3f)return true;
else return false;
}
int calc() {
int ret=0;
int maxx = 0x3f3f3f3f;
for(int i=T;i!=S;i=edge[pre[i]].u)
maxx=min(edge[pre[i]].flow,maxx);
for(int i=T;i!=S;i=edge[pre[i]].u) {
edge[pre[i]].flow-=maxx;
edge[pre[i]^1].flow+=maxx;
ret+=maxx*edge[pre[i]].cost;
}
return ret;
}
int ans=0;
void mcfc() {
while(spfa())
ans+=calc();
}
int main() {
m=read(),n=read();
T=n+n*m+1;
for(int i=1;i<=n;++i) {
for(int j=1;j<=m;++j) {
peo[i][j]=read();
}
}
for(int i=1;i<=n;++i) add_edge(S,i,1,0);
for(int i=1;i<=m;++i) {
for(int j=1;j<=n;++j) {
add_edge(n*i+j,T,1,0);
}
}
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j)
for(int k=1;k<=n;++k)
add_edge(i,n*j+k,1,k*peo[i][j]);
mcfc();
printf("%.2lf\n",(double)ans/n);
return 0;
}