POJ 3686 The Windy's 最小權值匹配

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點擊開啟連結The Windy‘s

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3788		Accepted: 1630

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

有n個任務需要m台機器去做，每台機器一次只能做一個任務，而且每個任務必須經過這m台機器才能完成，先給你每個任務在每台機器的時間，問最少的平均時間是多少。 完成所有的任務總時間是實際時間+等待時間，設完成1,2,3....n的時間分別為t1,t2,t3...tn，那麼總時間就是t=t1+(t1+t2)+(t1+t2+t3)+......+(t1+t2+...+tn)=t1*n+t2*(n-1)+t3*(n-2)+tn 第k個任務在倒數第j台機器處理tk*j。 每個機器可以處理n個任務，將每個機器拆成n個點， 1~n分別代表某個任務在這個機器上倒數第幾個被加工的，對於每個任務i，機器j中拆的每個點k，串連一條-map[i][j]*k權值的邊。

//36044K579MS#include<stdio.h>#include<string.h>#include<algorithm>#define M 3007#define inf 0x3f3f3fusing namespace std;int g[M][M],map[M][M];int lx[M],ly[M];int slack[M],match[M];bool visx[M],visy[M];int n,m;void build(){    for(int i=1; i<=n; i++)        for(int j=1; j<=m; j++)            for(int k=1; k<=n; k++)                g[i][(j-1)*n+k]=-map[i][j]*k;    m=n*m;}bool dfs(int cur){    visx[cur]=true;    for(int y=1; y<=m; y++)    {        if(!visy[y]&&lx[cur]+ly[y]==g[cur][y])        {            visy[y]=true;            if(match[y]==-1||dfs(match[y]))            {                match[y]=cur;                return true;            }        }    }    return false;}int KM(){    memset(match,-1,sizeof(match));    memset(ly,0,sizeof(ly));    for(int i=1; i<=n; i++)    {        lx[i]=-inf;        for(int j=1; j<=m; j++)            lx[i]=max(lx[i],g[i][j]);    }    for(int x=1; x<=n; x++)    {        while(true)        {            memset(visx,false,sizeof(visx));            memset(visy,false,sizeof(visy));            if(dfs(x))break;            int d=inf;            for(int j=1; j<=n; j++)                if(visx[j])                {                    for(int k=1; k<=m; k++)                    {                        if(!visy[k]&&d>lx[j]+ly[k]-g[j][k])                        {                            d=lx[j]+ly[k]-g[j][k];                        }                    }                }            for(int i=1; i<=n; i++)                if(visx[i])                    lx[i]-=d;            for(int i=1; i<=m; i++)                if(visy[i])                    ly[i]+=d;        }    }    int result=0;    for(int i=1; i<=m; i++)    {        if(match[i]!=-1&&g[match[i]][i]!=-inf)            result+=g[match[i]][i];    }    return result;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        memset(g,0,sizeof(g));        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                scanf("%d",&map[i][j]);        build();        int ans=KM();        printf("%.6f\n",-1.0*(double)ans/n);    }    return 0;}