poj3686 The Windy's(拆点,费用流,好题)

题目链接

The Windy's

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5031		Accepted: 2112

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

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

题解：

以前没见过的一种拆点方式,挑战程序设计上P243上的一道题。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAXN=4000;
const int MAXM=MAXN*MAXN;
const int INF = 0x3f3f3f3f;
int g[55][55];

struct Edge
{
int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t)
{
queue<int> q;
for(int i = 0;i < N;i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//返回的是最大流，cost存的是最小费用
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 != -1;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 != -1;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 main()
{
int cas;
scanf("%d",&cas);
while(cas--)
{
int n,m;
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",&g[i][j]);
int start=0,end=n+n*m+1;
init(end+1);
for(int i=1;i<=n;i++) addedge(start,i,1,0);
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++)
addedge(n+m*(j-1)+i,end,1,0);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
for(int k=1;k<=n;k++)
addedge(i,n+m*(k-1)+j,1,g[i][j]*k);
int cost=0;
minCostMaxflow(start,end,cost);
double ans=1.0*cost/n;
printf("%.6f\n",ans);
}
return 0;
}