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The Windy‘s

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5477		Accepted: 2285

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 Z_ij 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 Z_ij (1 ≤ Z_ij ≤ 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

3

3 4
100 100 100 1
99 99 99 1
98 98 98 1

3 4
1 100 100 100
99 1 99 99
98 98 1 98

3 4
1 100 100 100
1 99 99 99
98 1 98 98

Sample Output

2.000000
1.000000
1.333333

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778 題目鏈接：http://poj.org/problem?id=3686 題意：有n個玩具m個工廠，一個工廠同時只能生產一個玩具。問平均每個玩具的生產時間。 思路：現在有n個玩具，m個工廠，但是一個工廠可以生產多個玩具。當一個工廠按順序生產a1，a2，a3...，an玩具，每個玩具的生產時間為Za1，Za1+Za2，Za1+Za2+Za3，...，Za1+Za2+Za3+...+Zan。T=n*Za1+(n-1)*Za2+(n-2)*Za3+...+Zan。這個式子也可以理解為，多個只能生產1個玩具的工廠，但是時間為1~n倍。
代碼：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define PI acos(-1.0)
const int maxn=3e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;
struct edge
{
    int from,to;
    int cap,cost;
    int rev;
};
int NN;
vector<edge>G[maxn];
int h[maxn];
///頂點的勢，取h(u)=(s到u的最短距離)，邊e=(u,v)的長度變成d`(e)=d(e)+h(u)-h(v)>=0
int dist[maxn];
int prevv[maxn],preve[maxn];///前驅結點和對應的邊
void addedge(int u,int v,int cap,int cost)
{
    edge e;
    e.from=u,e.to=v,e.cap=cap,e.cost=cost,e.rev=G[v].size();
    G[u].push_back(e);
    e.from=v,e.to=u,e.cap=0,e.cost=-cost,e.rev=G[u].size()-1;
    G[v].push_back(e);
}
int min_cost_flow(int s,int t,int f)
{
    int res=0;
    fill(h,h+NN,0);
    while(f>0)
    {
        priority_queue<P,vector<P>,greater<P> >q;
        fill(dist+1,dist+NN,inf);
        dist[s]=0;
        q.push(P(dist[s],s));
        while(!q.empty())
        {
            P p=q.top();
            q.pop();
            int u=p.second;
            if(dist[u]<p.first) continue;
            for(int i=0; i<G[u].size(); i++)
            {
                edge e=G[u][i];
                if(e.cap>0&&dist[e.to]>dist[u]+e.cost+h[u]-h[e.to])
                {
                    dist[e.to]=dist[u]+e.cost+h[u]-h[e.to];
                    prevv[e.to]=u;
                    preve[e.to]=i;
                    q.push(P(dist[e.to],e.to));
                }
            }
        }
        if(dist[t]==inf) return res;
        for(int i=0; i<NN; i++) h[i]+=dist[i];
        int d=f;
        for(int i=t; i!=s; i=prevv[i])
            d=min(d,G[prevv[i]][preve[i]].cap);
        f-=d;
        res+=d*h[t];
        for(int i=t; i!=s; i=prevv[i])
        {
            //cout<<i<<" ";
            edge &e=G[prevv[i]][preve[i]];
            e.cap-=d;
            G[i][e.rev].cap+=d;
        }
        //cout<<s<<endl;
    }
    return res;
}
int z[100][100];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int s=0,t=n+n*m+1;
        NN=t+1;
        for(int i=1; i<=n; i++)
        {
            addedge(s,i,1,0);
            for(int j=1; j<=m; j++)
            {
                scanf("%d",&z[i][j]);
                for(int k=1; k<=n; k++)
                    addedge(i,n+(j-1)*n+k,1,k*z[i][j]);
            }
        }
        for(int j=1; j<=m; j++)
        {
            for(int k=1; k<=n; k++)
                addedge(n+(j-1)*n+k,t,1,0);
        }
        printf("%.6f\n",min_cost_flow(s,t,inf)*1.0/n);
        for(int i=0; i<NN; i++) G[i].clear();
    }
    return 0;
}

最小費用最大流

POJ 3686.The Windy's 最小費用最大流