T - Taxi Cab Scheme
Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2060
Appoint description:
System Crawler (2014-08-22)
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 今天我又把题目粘过来了， 原因很简单， 我又读错题了 TAT 大意：有n个点，给你一个起始时间，从一个点到另一个点需要的时间是横纵坐标差的绝对值之和， 问需要派出多少cab分析：最小路径覆盖 = n - 最大匹配 代码：
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <vector>
5 #include <cmath>
6 using namespace std;
7
8 const int maxn = 505;
9 const int INF = 1000000000;
10
11 int n;
12 int vis[maxn];
13 int Link[maxn];
14 vector<int> G[maxn];
15 bool Find(int u) {
16 for(int i = 0; i < G[u].size(); i++) {
17 int v = G[u][i];
18 if(!vis[v]) {
19 vis[v] = 1;
20 if(Link[v] == - 1 || Find(Link[v])) {
21 Link[v] = u;
22 return true;
23 }
24 }
25 }
26 return false;
27 }
28
29 int solve() {
30 int cnt = 0;
31 memset(Link, -1, sizeof(Link));
32 for(int i = 1; i <= n; i++) {
33 if(G[i].size()) {
34 memset(vis, 0, sizeof(vis));
35 if(Find(i)) cnt++;
36 }
37 }
38 return cnt;
39 }
40
41 int a[maxn], b[maxn], c[maxn], d[maxn], t[maxn];
42 bool check(int i, int j) {
43 if(t[j] - ((fabs(c[i] - a[i]) + fabs(d[i] - b[i]) + t[i]) + fabs(a[j] - c[i]) + fabs(b[j] - d[i])) >= 1) {
44 return true;
45 }
46 return false;
47 }
48
49 int main() {
50 int tt;
51 int x, y;
52 scanf("%d",&tt);
53 while(tt--) {
54 scanf("%d",&n);
55 for(int i = 1; i <= n; i++) {
56 G[i].clear();
57 scanf("%d:%d",&x,&y);
58 t[i] = x * 60 + y;
59 scanf("%d %d %d %d",&a[i], &b[i], &c[i], &d[i]);
60 }
61 for(int i = 1; i <= n; i++) {
62 for(int j = 1; j <= n; j++) {
63 if(i == j) continue;
64 if(check(i, j)) {
65 G[i].push_back(j);
66 }
67 }
68 }
69 printf("%d\n", n - solve());
70 }
71 return 0;
72 }