poj 2060(最小路径覆盖)

Taxi Cab Scheme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4207		Accepted: 1750

Description

Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up the customers calling to get a cab as soon as possible,there is also a need to schedule all the taxi rides which have been booked in advance.Given a list of all booked taxi rides for the next day, you want to minimise the number of cabs needed to carry out all of the rides. 
For the sake of simplicity, we model a city as a rectangular grid. An address in the city is denoted by two integers: the street and avenue number. The time needed to get from the address a, b to c, d by taxi is |a - c| + |b - d| minutes. A cab may carry out a booked ride if it is its first ride of the day, or if it can get to the source address of the new ride from its latest,at least one minute before the new ride's scheduled departure. Note that some rides may end after midnight.

Input

On the first line of the input is a single positive integer N, telling the number of test scenarios to follow. Each scenario begins with a line containing an integer M, 0 < M < 500, being the number of booked taxi rides. The following M lines contain the rides. Each ride is described by a departure time on the format hh:mm (ranging from 00:00 to 23:59), two integers a b that are the coordinates of the source address and two integers c d that are the coordinates of the destination address. All coordinates are at least 0 and strictly smaller than 200. The booked rides in each scenario are sorted in order of increasing departure time.

Output

For each scenario, output one line containing the minimum number of cabs required to carry out all the booked taxi rides.

Sample Input

2
2
08:00 10 11 9 16
08:07 9 16 10 11
2
08:00 10 11 9 16
08:06 9 16 10 11

Sample Output

1
2

Source

Northwestern Europe 2004

题目： http://poj.org/problem?id=2060

分析：跟上题类似，此题也是最小路径覆盖，将每个客人的需求分为开始和结束两点，增加源和汇，将源与每个要求的结束点连上容量为1 的边，每个开始点与汇连上容量为1 的边，然后枚举结束点i，和开始点j，如果i能够赶到j的话，就连上一条边，然后答案为m-最大流。。。这题用二分图应该会快些吧～～～

代码：

#include<cstdio>
using namespace std;
const int mm=555555;
const int mn=1111;
const int oo=1000000000;
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];
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;
}