http://poj.org/problem?id=2060
Taxi Cab Scheme
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 5459 | Accepted: 2286 |
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
【题解】：
题意：有n个任务：开始时间、起始地点、终止地点。每个地点可以派出一辆出租车，如果出租车完成任务i后还可以到达任务j，那么它可以继续执行任务j。现在问最少可以排除多少辆出租车?
算法：1、最小路径覆盖
2、在无圈有向图中：最小路径覆盖＝｜Ｐ｜－最大匹配数。
3、建图：如果任务i和任务j可以由一辆出租车共同执行，则将i和j连线。满足的条件如下：
任务i的开始时间+任务i的完成时间+从任务i的目的地到达任务j的起始地点所花费的时间<任务j的开始时间
算法：1、最小路径覆盖
2、在无圈有向图中：最小路径覆盖＝｜Ｐ｜－最大匹配数。
3、建图：如果任务i和任务j可以由一辆出租车共同执行，则将i和j连线。满足的条件如下：
任务i的开始时间+任务i的完成时间+从任务i的目的地到达任务j的起始地点所花费的时间<任务j的开始时间
【code】：
1 /** 2 Judge Status:Accepted Memory:1888K 3 Time:157MS Language:G++ 4 Code Lenght:1687B Author:cj 5 */ 6 #include<iostream> 7 #include<stdio.h> 8 #include<string.h> 9 #include<math.h> 10 #include<algorithm> 11 12 #define N 550 13 using namespace std; 14 15 struct Nod 16 { 17 int time,sx,sy,ex,ey; //时间以及始末坐标 18 }node[N]; 19 20 int n,m; 21 int map[N][N]; 22 int cx[N],cy[N],mark[N]; 23 24 int abs(int x){return x>0?x:-x;} 25 26 int path(int u) 27 { 28 int j; 29 for(j=1;j<=n;j++) 30 { 31 if(map[u][j]&&!mark[j]) 32 { 33 mark[j]=1; 34 if(cy[j]==-1||path(cy[j])) 35 { 36 cx[u] = j; 37 cy[j] = u; 38 return 1; 39 } 40 } 41 } 42 return 0; 43 } 44 45 int maxMatch() //求最大节点覆盖 46 { 47 memset(cx,-1,sizeof(cx)); 48 memset(cy,-1,sizeof(cy)); 49 int i; 50 int res = 0; 51 for(i=1;i<=n;i++) 52 { 53 if(cx[i]==-1) 54 { 55 memset(mark,0,sizeof(mark)); 56 res+=path(i); 57 } 58 } 59 return res; 60 } 61 62 int main() 63 { 64 int t; 65 scanf("%d",&t); 66 while(t--) 67 { 68 scanf("%d",&n); 69 int i; 70 for(i=1;i<=n;i++) 71 { 72 int h,m; 73 scanf("%d:%d%d%d%d%d",&h,&m,&node[i].sx,&node[i].sy,&node[i].ex,&node[i].ey); 74 node[i].time = h*60+m; //二维时间化一维 75 } 76 memset(map,0,sizeof(map)); 77 int j; 78 for(i=1;i<=n;i++) 79 { 80 for(j=1;j<=n;j++) 81 { 82 if(i!=j&&(abs(node[i].sx-node[i].ex)+abs(node[i].sy-node[i].ey)+node[i].time+ 83 abs(node[i].ex-node[j].sx)+abs(node[i].ey-node[j].sy)<node[j].time)) 84 { //任务i的开始时间+任务i的完成时间+从任务i的目的地到达任务j的起始地点所花费的时间<任务j的开始时间 85 map[i][j]=1; 86 } 87 } 88 } 89 printf("%d ",n-maxMatch()); //节点数 - 最大节点覆盖 = 最小路径覆盖 90 } 91 return 0; 92 }