POJ 3686 The Windy's

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拆點+最小費用最大流:

將工廠拆點,表示再第j個工廠倒數第k個生產則第i個玩具到(j,k)工廠的費用為 k*t[i][j]

The Windy‘s Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4066   Accepted: 1746 Description The Windy‘s is a world famous toy factory that owns M top-class workshop to make toys. This year the manager receives N orders for toys. The manager knows that every order will take different amount of hours in different workshops. More precisely, the i-th order will take Zij hours if the toys are making in the j-th workshop. Moreover, each order‘s work must be wholly completed in the same workshop. And a workshop can not switch to another order until it has finished the previous one. The switch does not cost any time. The manager wants to minimize the average of the finishing time of the N orders. Can you help him? Input The first line of input is the number of test case. The first line of each test case contains two integers, N and M (1 ≤ N,M ≤ 50). The next N lines each contain M integers, describing the matrix Zij (1 ≤ Zij ≤ 100,000) There is a blank line before each test case. Output For each test case output the answer on a single line. The result should be rounded to six decimal places. Sample Input 33 4100 100 100 199 99 99 198 98 98 13 41 100 100 10099 1 99 9998 98 1 983 41 100 100 1001 99 99 9998 1 98 98 Sample Output 2.0000001.0000001.333333 Source POJ Founder Monthly Contest – 2008.08.31, windy7926778

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn=5000;const int INF=0x3f3f3f3f;struct Edge{    int to,next,cap,flow,cost;} edge[1500000];int Adj[maxn],Size,N;void init(){    Size=0;    memset(Adj,-1,sizeof(Adj));}void addedge(int u,int v,int cap,int cost){    edge[Size].to=v;    edge[Size].next=Adj[u];    edge[Size].cost=cost;    edge[Size].cap=cap;    edge[Size].flow=0;    Adj[u]=Size++;}void Add_Edge(int u,int v,int cap,int cost){    addedge(u,v,cap,cost);    addedge(v,u,0,-cost);}int dist[maxn],vis[maxn],pre[maxn];bool spfa(int s,int t){    queue<int> q;    for(int i=0; i<N; i++)    {        dist[i]=INF;        vis[i]=false;        pre[i]=-1;    }    dist[s]=0;    vis[s]=true;    q.push(s);    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=false;        for(int i=Adj[u]; ~i; i=edge[i].next)        {            int v=edge[i].to;            if(edge[i].cap>edge[i].flow&&                    dist[v]>dist[u]+edge[i].cost)            {                dist[v]=dist[u]+edge[i].cost;                pre[v]=i;                if(!vis[v])                {                    vis[v]=true;                    q.push(v);                }            }        }    }    if(pre[t]==-1) return false;    return true;}int MinCostMaxFlow(int s,int t,int& cost){    int flow=0;    cost=0;    while(spfa(s,t))    {        int Min=INF;        for(int i=pre[t]; ~i; i=pre[edge[i^1].to])        {            if(Min>edge[i].cap-edge[i].flow)                Min=edge[i].cap-edge[i].flow;        }        for(int i=pre[t]; ~i; i=pre[edge[i^1].to])        {            edge[i].flow+=Min;            edge[i^1].flow-=Min;            cost+=edge[i].cost*Min;        }        flow+=Min;    }    return flow;}int n,m;int tv[60][60];int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        init();        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                scanf("%d",&tv[i][j]);        for(int i=1; i<=n; i++) ///i號玩具        {            Add_Edge(0,i,1,0);            for(int j=1; j<=m; j++) ///m類工廠                for(int k=1; k<=n; k++) ///倒數第K個加工                {                    Add_Edge(i,j*n+k,1,tv[i][j]*k);                    if(i==1)    Add_Edge(j*n+k,(m+1)*n+1,1,0);                }        }        N=(m+1)*n+2;        int cost,flow;        flow=MinCostMaxFlow(0,N-1,cost);        printf("%.6lf\n",1.*cost/flow);    }    return 0;}

POJ 3686 The Windy's