1070: [SCOI2007]修車

/*
一開始以為是個貪心 發現自己太naive了將每個技術工人拆成n個點，一共拆n*m個，第i個表示倒數第i次修車。 
讓每輛車向拆出來的點連邊，費用為tmp[i][j]*k，i是技工，j是車，k是拆出來的第幾個點，
這樣設置費用的原因是j是i倒數第k個修的，那么i修的后k個車都要等倒數第k個車。
然后跑最小費用最大流就可以了
*/#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;inline int read()
{char c=getchar();int num=0;for(;!isdigit(c);c=getchar());for(;isdigit(c);c=getchar())num=num*10+c-'0';return num;
}const int N=65;
const int M=2e5+5;int m,n,S,T;
int tim[N][N];
long long ans;int head[M],num_edge;
struct Edge
{int v,nxt,flow,cost;
}edge[M];inline void add_edge(int u,int v,int flow,int cost)
{edge[++num_edge].v=v;edge[num_edge].flow=flow;edge[num_edge].cost=cost;edge[num_edge].nxt=head[u];head[u]=num_edge;edge[++num_edge].v=u;edge[num_edge].flow=0;edge[num_edge].cost=-cost;edge[num_edge].nxt=head[v];head[v]=num_edge;
}int dis[M];
queue<int> que;
bool inque[M];
bool spfa()
{memset(dis,0x3f,sizeof(dis));que.push(S),dis[S]=0;int now;while(!que.empty()){now=que.front(),que.pop();for(int i=head[now],v;i;i=edge[i].nxt){if(edge[i].flow==0)continue;v=edge[i].v;if(dis[v]>dis[now]+edge[i].cost){dis[v]=dis[now]+edge[i].cost;if(!inque[v]){que.push(v);inque[v]=1;}}}inque[now]=0;}return dis[T]!=0x3f3f3f3f;
}int vis[M],visf;
int dfs(int u,int flow)
{if(u==T||!flow)return flow;int outflow=0;vis[u]=visf;for(int i=head[u],v,tmp;i;i=edge[i].nxt){if(!edge[i].flow)continue;v=edge[i].v;if(vis[v]!=visf&&dis[v]==dis[u]+edge[i].cost){tmp=dfs(v,min(flow,edge[i].flow));if(!tmp)continue;ans+=1ll*tmp*edge[i].cost;edge[i].flow-=tmp;edge[i^1].flow+=tmp;flow-=tmp;outflow+=tmp;if(!flow){vis[u]=0;return outflow;}}}vis[u]=0;dis[u]=0x7fffffff;return outflow;
}int main()
{num_edge=1;m=read(),n=read();T=n+n*m+1;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)tim[j][i]=read();for(int i=1;i<=n;++i)add_edge(S,i,1,0);for(int i=1,p;i<=m;++i){for(int j=1;j<=n;++j){p=i*n+j;add_edge(p,T,1,0);for(int c=1;c<=n;++c)add_edge(c,p,1,tim[i][c]*j);}}while(spfa()){++visf;dfs(S,0x7fffffff);}printf("%.2lf",1.0*ans/n);return 0;
}

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