UVA1201 Taxi Cab Scheme(最小路径覆盖)

2018 年 08 月 30 日 • 阅读: 1001 • 图论阅读设置

Taxi Cab Scheme

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 7529 Accepted: 3083


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.


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.


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

Sample Input

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

Sample Output



Northwestern Europe 2004







本题中,“时间”是一个天然的序,因此可以构图如下:每个客人作为一个结点,如果同一个出租车在接完客人u后来得及接客人v,那么连边$u \rightarrow v$,不难发现,这个图是一个DAG,并且它的最小路径覆盖就是答案。




#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 510;
const int maxm = 250010;
using  namespace std;

struct Edge
    int to, next;
int head[maxn], tot;
int linker[maxn];
bool used[maxn];

struct Node
    int hh, mm;
    int x1, y1, x2, y2;

void init()
    tot = 0;
    memset(head, -1, sizeof(head));

void addedge(int u, int v)
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;

bool dfs(int u)
    for (int i = head[u]; i != -1; i = edge[i].next)
        int v = edge[i].to;
        if (!used[v])
            used[v] = true;
            if (linker[v] == -1 || dfs(linker[v]))
                linker[v] = u;
                return true;
    return false;

int hungary(int n) // 最大匹配
    int res = 0;
    memset(linker, -1, sizeof(linker));
    for (int u = 0; u < n; ++u)
        memset(used, 0, sizeof(used));
        if (dfs(u))
    return res;

int time_hours(Node a)
    return a.hh * 60 +;

int time(Node a, Node b)
    return abs(a.x1 - a.x2) + abs(a.y1 - a.y2) + abs(a.x2 - b.x1) + abs(a.y2 - b.y1);

int main()
//    freopen("/Users/taifu/Codes/NOW/", "r", stdin);
//    freopen("/Users/taifu/Codes/NOW/data.out","w",stdout);
    int T, m;
    scanf("%d", &T);
    while (T--)
        scanf("%d", &m);
        for (int i = 0; i < m; ++i)
            scanf("%d:%d %d%d%d%d", &node[i].hh, &node[i].mm, &node[i].x1, &node[i].y1, &node[i].x2, &node[i].y2);
        for (int i = 0; i < m; ++i)
            for (int j = i + 1; j < m; ++j)
                if (time_hours(node[i]) + time(node[i], node[j]) < time_hours(node[j])) // 符合条件,连边
                    addedge(i, j);
        int ans = m - hungary(m); // 点数-最大匹配数
        printf("%d\n", ans);
    return 0;


The end.
2018-08-30 星期四
最后编辑于: 2018 年 09 月 01 日