Hdu 4494 Teamwork（最小费用流）

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=4494

思路：每种属性人互不干扰，跑m次费用流，结果累加。超级源点0，超级汇点2*n-1。将每个点拆成两个点一个为自己，另一个表示可以提供给别人。源点向每个表示自己的点连一条容量为INF，费用为1的边，表示起点有无数人每选择一人需花费1。对于每个拆出来的点，源点向其连一条容量为kind[i]（第i中需要人数），费用为0的边。表示当前这个点最多可以提供kind[i]的人，费用为0。对于第i个任务，若worker[i].b+worker[i].p+dist<=worker[j].b，则将i拆出来点向表示j自己的点连边，表示可以提供worker。

#include<cstdio>#include<queue>#include<cmath>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#define debuusing namespace std;const int maxn=500;const int INF=0x3f3f3f3f;typedef long long LL;struct Edge{    int from,to,cap,flow,cost;    Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w) {}};struct Node{    int kind[6];    int b,p,x,y;};int s,t,stx,sty;Node worker[maxn];vector<Edge> edges;vector<int> G[maxn];int n,m,p[maxn],a[maxn];int inq[maxn],d[maxn];LL cost;void init(int n){    edges.clear();    for(int i=0; i<=n; i++) G[i].clear();}void addedges(int from,int to,int cap,int cost){    edges.push_back(Edge(from,to,cap,0,cost));    edges.push_back(Edge(to,from,0,0,-cost));    int tot=edges.size();    G[from].push_back(tot-2);    G[to].push_back(tot-1);}bool BF(int s,int t,int& flow,LL& cost){    //cout<<n<<" "<<m<<endl;    //cout<<"flag"<<endl;    for(int i=0; i<=2*n; i++) d[i]=INF;    memset(inq,0,sizeof(inq));    d[s]=0,inq[s]=1,p[s]=0;    queue<int> q;    q.push(s),a[s]=INF;    while(!q.empty())    {        int u=q.front();        q.pop(),inq[u]=0;        for(int i=0; i<G[u].size(); i++)        {            Edge& e=edges[G[u][i]];            if(e.cap>e.flow&&d[e.to]>d[u]+e.cost)            {                d[e.to]=d[u]+e.cost;                p[e.to]=G[u][i];                a[e.to]=min(a[u],e.cap-e.flow);                if(!inq[e.to])                {                    q.push(e.to);                    inq[e.to]=1;                }            }        }    }    if(d[t]==INF) return false;    flow+=a[t];    cost+=(LL)d[t]*(LL)a[t];    for(int u=t; u!=s; u=edges[p[u]].from)    {        edges[p[u]].flow+=a[t];        edges[p[u]^1].flow-=a[t];    }    return true;}int mincostmaxflow(int s,int t,LL& cost){    int flow=0;    cost=0;    while(BF(s,t,flow,cost));    return cost;}double dist(Node a,Node b){    return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y));}void make(int x){    init(2*n);    for(int i=1; i<=n-1; i++)    {        addedges(s,i,INF,1);        addedges(i,t,worker[i].kind[x],0);        addedges(s,i+n-1,worker[i].kind[x],0);        //cout<<s<<" "<<i<<" "<<INF<<" "<<1<<endl;        //cout<<i<<" "<<t<<" "<<worker[i].kind[x]<<" "<<0<<endl;        //cout<<s<<" "<<i+n-1<<" "<<worker[i].kind[x]<<" "<<0<<endl;        for(int j=1; j<=n-1; j++)        {            if(i==j) continue;            double dis=dist(worker[i],worker[j]);            if(worker[i].b+worker[i].p+dis<=worker[j].b)            {                //cout<<i+n-1<<" "<<j<<" "<<INF<<" "<<0<<endl;                addedges(i+n-1,j,INF,0);            }        }    }    //cout<<"flag"<<endl;}void solve(){    int ans=0;    for(int i=1; i<=m; i++)    {        make(i);        //cout<<s<<" "<<t<<endl;        ans+=mincostmaxflow(s,t,cost);    }    printf("%d\n",ans);}int main(){#ifdef debug    freopen("in.in","r",stdin);#endif // debug    int cas;    scanf("%d",&cas);    while(cas--)    {        scanf("%d%d",&n,&m);        scanf("%d%d",&stx,&sty);        s=0,t=2*(n-1)+1;        //cout<<s<<" "<<t<<endl;        for(int i=1; i<=n-1; i++)        {            scanf("%d%d%d%d",&worker[i].x,&worker[i].y,&worker[i].b,&worker[i].p);            for(int j=1; j<=m; j++)                scanf("%d",&worker[i].kind[j]);        }        solve();    }    return 0;}