poj2060最小路径覆盖

俄罗斯套娃，如果i能装到j里面，则从i到j建一条有向边，问最后能看到几个娃娃，即需要多少条路径能够覆盖到所有节点。

#include <cstdio>
using namespace std;
const int oo=1e9;
const int mm=111111;
const int mn=999;
int node,src,dest,edge;
int ver[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn];
int sx[mn],sy[mn],ex[mn],ey[mn],st[mn],et[mn];
struct node
{
    int l,h,w;
}p[mn];
inline int min(int a,int b)
{
    return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0;i<node;++i)head[i]=-1;
    edge=0;
}
inline void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0;i<node;++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0;l<r;++l)
        for(i=head[u=q[l]];i>=0;i=next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp;i>=0;i=next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0;i<node;++i)work[i]=head[i];
        while(delta=Dinic_dfs(src,oo))ret+=delta;
    }
    return ret;
}
inline int get(int a)
{
    return a<0?-a:a;
}

int main()
{
    int m,s,i,j,n,t;
    char c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        prepare(n+n+2,0,n+n+1);
        for(i=1;i<=n;++i)
        {
            scanf("%d%c%d%d%d%d%d",&m,&c,&s,&sx[i],&sy[i],&ex[i],&ey[i]);
            et[i]=(st[i]=m*60+s)+get(sx[i]-ex[i])+get(sy[i]-ey[i]);
        }
        for(i=1;i<=n;++i)
        {
            addedge(src,i,1),addedge(i+n,dest,1);
            for(j=i+1;j<=n;++j)
                if(et[i]+get(ex[i]-sx[j])+get(ey[i]-sy[j])<st[j])addedge(i,n+j,1);
        }
        printf("%d\n",n-Dinic_flow());
    }
    return 0;
}