poj 2060 Taxi Cab Scheme (二分匹配)

Taxi Cab Scheme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5710		Accepted: 2393

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

 

二分匹配：

弄清楚题意比较重要：

出租车公司有n个预约, 每个预约有时间和地点, 地点分布在二维整数坐标系上, 地点之间的行驶时间为两点间的曼哈顿距离（|x1 - x2| + |y1 - y2|）。一辆车可以在运完一个乘客后运另一个乘客, 条件是此车要在预约开始前一分钟之前到达出发地, 问最少需要几辆车搞定所有预约。(摘自http://blog.sina.com.cn/s/blog_6635898a0100m54w.html)

 

我就是没弄清题意WA了好几次。弄清提议后开始构图，求最小边覆盖，还有就是这是有向图有所以构单边。

//692K    79MS    C++    1333B    2014-06-05 11:26:44
#include<iostream>
#include<vector>
#define N 505
using namespace std;
struct node{
    int a,b,c,d;
    int time;
}p[N];
vector<int>V[N];
int match[N];
int vis[N];
int n;
int dfs(int u)
{
    for(int i=0;i<V[u].size();i++){
        int v=V[u][i];
        if(!vis[v]){
            vis[v]=1;
            if(match[v]==-1 || dfs(match[v])){
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    int ret=0;
    memset(match,-1,sizeof(match));
    for(int i=0;i<n;i++){
        memset(vis,0,sizeof(vis));
        ret+=dfs(i);
    }
    return ret;
}
int main(void)
{
    int t;
    int time[N];
    int dis[N];
    int h,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<=n;i++) V[i].clear();
        for(int i=0;i<n;i++){
            scanf("%d:%d %d%d%d%d",&h,&m,&p[i].a,&p[i].b,&p[i].c,&p[i].d);
            p[i].time=abs(p[i].a-p[i].c)+abs(p[i].b-p[i].d);
            time[i]=h*60+m;
            for(int j=0;j<i;j++)
                if(time[i]-time[j]>p[j].time+abs(p[i].a-p[j].c)+abs(p[i].b-p[j].d)){
                    //V[i].push_back(j);
                    V[j].push_back(i);
                }
        }
        printf("%d\n",n-hungary());
    }
    return 0;
}