POJ3686 The Windy's【网络流】

题意：有m个工厂，n个订单，问平均完成时间

思路：假设一个工厂依次完成n1,n2,n3，时间为3*t1 + 2*t2 + t3，一个任务在倒数第k个完成，时间为k*t。

把一个工厂分为n个时间点，枚举每个任务完成的时间，连到相应的时间点，每个工厂的时间点汇集在起点，容量为1，每个任务汇集在终点，容量为1，费用都为0

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;const int maxn = 2600;struct edge{int to,cf,w,rev;};int d[maxn],fa[maxn],inq[maxn],pr[maxn];vector<struct edge> v[maxn];void init(int x){for(int i = 0; i <= x; i++)v[i].clear();}void add(int from,int to,int cap,int cost){v[from].push_back((edge){to,cap,cost,v[to].size()});v[to].push_back((edge){from,0,-cost,v[from].size()-1});}int mincost(int s,int t){queue<int> q;while(!q.empty())q.pop();int c,f;c = f = 0;while(1){memset(fa,-1,sizeof fa);memset(d,0x3f,sizeof d);d[s] = 0;memset(inq,0,sizeof inq);q.push(s);while(!q.empty()){int x = q.front();q.pop();inq[x] = 0;for(int i = 0; i < v[x].size(); i++){edge e = v[x][i];int y = e.to;if(e.cf && d[y] > d[x] + e.w){d[y] = d[x] + e.w;fa[y] = x;pr[y] = i;if(!inq[y]){inq[y] = 1;q.push(y);}}}}if(d[t] == 0x3f3f3f3f)break;int gen = 0x3f3f3f3f;for(int a = t; a != s; a = fa[a])gen = min(gen,v[fa[a]][pr[a]].cf);for(int a = t; a != s; a = fa[a]){edge &e = v[fa[a]][pr[a]];e.cf -= gen;v[a][e.rev].cf += gen;}f += gen;c += gen * d[t];}return c;//f 最大流，c最小费用 }int main(void){int T,n,m,i,j,k;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);int s = 1;int t = 2 + m*n + n;init(t);for(i = 1; i <= n; i++){for(j = 1; j <= m; j++){int tp;scanf("%d",&tp);for(k = 1; k <= n; k++){add(1+(j-1)*n+k,1+n*m+i,1,k * tp);}}}for(i = 1; i <= m; i++){for(j = 1; j <= n; j++){add(1,1+(i-1)*n+j,1,0);}}for(i = 1; i <= n; i++)add(1+n*m+i,t,1,0);double ans = mincost(s,t) / (n*1.0);printf("%f\n",ans);}return 0;}

G++编译，double 输出要用%f