HDU 4494 Teamwork 費用流/網絡流

Problem Description Some locations in city A has been destroyed in the fierce battle. So the government decides to send some workers to repair these locations. There are m kinds of workers that were trained for different skills. Each location need some number of some kinds of workers and has a schedule that at what time can the repair begins, and the time cost of repair. Any job cannot begin until all the workers required arrived.

For example, location 1 needs 2 workers of type 1 and 3 workers of type 2, and the beginning time and time cost is 100 minute and 90 minute correspondingly, then 5 workers that satisfy the requirement should arrive before 100 minute, start working at 100 minute and get the job done at 190 minute. Notice that two different types of workers cannot replace each other, so with 3 workers of type 1 and only 2 workers of type 2, this job cannot be done.

Workers can go from one location to another after their jobs are done. You can take the Euclidean distance between locations as the time workers need to travel between them. Each worker should be sent from a depot initially at 0 minute. Now your task is to determine the minimum number of workers needed to be sent from depot so that all the jobs can be done.  

Input There are multiple test cases, the integer on the first line T (T<25) indicates the number of test cases.

Each test case begins with two integers n (<=150), the number of location(including the depot) and m(<=5), the number of different skills.

The next line gives two integers x 0, y 0 indicates the coordinate of depot.

Then follows n - 1 lines begins with 4 integer numbers: x i, y i, b i(b i>0), p i(p i>0), (x i, y i) gives the coordinate of the i-th location, bi gives the beginning time and pi gives the time cost. The rest of the line gives m non-negative integers v 1, v 2, ..., v m, of which the i-th number indicates the the number of workers of type i needed (for all v i, 0<=v i<10, each location at least requires one worker).

All integers are less than 1000000 (10 6).  

Output For each test cases output one line, the minimum workers to be sent. It is guaranteed that there's always a feasible solution that all the jobs can be done.  

Sample Input

2 
4 1 
0 0 
0 1 1 1 3 
1 1 3 3 4 
1 0 10 1 5 
4 1 
0 0 
0 1 1 1 3 
1 1 3 3 4 
1 0 3 1 5 
        

Sample Output

5
9
        

一個地點拆成三個點，i、i’、i”

S -> i : 容量為type[i][k]，費用為1.

i -> i’ : 容量為type[i][k]，費用為0.

i’ -> T : 容量為type[i][k]，費用為0. S -> i” : 容量為type[i][k]，費用為0.

如果勞工可以在地點i工作完成後到達地點j開工，建邊i” -> j’ : 容量為v[i][k]，費用為0. 相當于j點的流量減小了v[i][k]。

最小路徑覆寫模型

可行流的最小流？  網絡流解法有待完善

#include<bits/stdc++.h>
#define mem(a,x) memset(a,x,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define MAX 100000

using namespace std;

#define ll int
#define inf 0x3f3f3f3f
//#define inf 999999999
//#define Inf 0x3FFFFFFFFFFFFFFFLL
#define N 500
#define M 50000

int x00,y00;
struct Point
{
   int x,y,b,p;
   int type[10];
}point[160],point0;

struct Edge
{
    ll to,cap,cost,nex;
    Edge() {}
    Edge(ll to,ll cap,ll cost,ll next):to(to),cap(cap),cost(cost),nex(next) {}
} edge[M];

ll head[N],top;
ll D[N],A[N],P[N],li[N];
bool inq[N];

double dis(Point p1,Point p2)
{
   return (sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)));
}

void add(ll from,ll to,ll cap,ll cost)
{
    edge[top]=Edge(to,cap,cost,head[from]);
    head[from]=top++;
    edge[top]=Edge(from,0,-cost,head[to]);
    head[to]=top++;
}

bool spfa(ll s,ll t,ll &flow,ll &cost)
{
    for(ll i=0; i<=t; i++) D[i]=inf;
    mem(inq,0);
    queue<ll>q;
    q.push(s);
    D[s]=0;
    A[s]=inf;
    while(!q.empty())
    {
        ll u=q.front();
        q.pop();
        inq[u]=0;
        for(ll i=head[u]; ~i; i=edge[i].nex)
        {
            Edge &e=edge[i];
            if(e.cap && D[e.to]>D[u]+e.cost)
            {
                D[e.to]=D[u]+e.cost;
                P[e.to]=i;
                A[e.to]=min(A[u],e.cap);
                if(!inq[e.to])
                {
                    inq[e.to]=1;
                    q.push(e.to);
                }
            }
        }
    }
    if(D[t]==inf) return false;
    cost+=D[t]*A[t];
    flow+=A[t];
    ll u=t;
    while(u!=s)
    {
        edge[P[u]].cap-=A[t];
        edge[P[u]^1].cap+=A[t];
        u=edge[P[u]^1].to;
    }
    return true;
}

ll mcmf(ll s,ll t)
{
    ll flow=0, cost=0;
    while(spfa(s,t,flow,cost));
    return cost;
}

int S,T,n,m,cnt;

void init(int TYPE)
{
    mem(head,-1);
    cnt=0;
    top=0;
    S=0;
    T=3*n+1;
    lp(i,n)
    {
        //if(point[i].b >= dis(point0,point[i]))
        {
           add(S,i,point[i].type[TYPE],1);
           add(S,i+2*n,point[i].type[TYPE],0);
        }

        add(i,i+n,point[i].type[TYPE],0);
        add(i+n,T,point[i].type[TYPE],0);

    }
    lp(i,n)
      lp(j,n)
      {
          if(i==j) continue;
          if((dis(point[i],point[j])+point[i].b+point[i].p) <= point[j].b)
          {
              //cout<<"i="<<i<<" "<<j<<endl;
              add(i+2*n,j+n,point[i].type[TYPE],0);
          }
      }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    sc(t);
    while(t--)
    {
        int re=0;
        sc2(n,m);
        sc2(x00,y00);
        point0.x=x00;
        point0.y=y00;
        lp(i,n-1)
         {
            sc2(point[i].x,point[i].y);
            sc2(point[i].b,point[i].p);
            lp(j,m)
            {
               sc(point[i].type[j]);
            }
         }
        n--;
        lp(i,m)
        {
            init(i);
            re+=mcmf(S,T);
        }
        printf("%d\n",re);
    }
    return 0;
}