L2-013. 红色警报-PAT团体程序设计天梯赛GPLT(图的连通分量个数统计)

战争中保持各个城市间的连通性非常重要。本题要求你编写一个报警程序,当失去一个城市导致国家被分裂为多个无法连通的区域时,就发出红色警报。注意:若该国本来就不完全连通,是分裂的k个区域,而失去一个城市并不改变其他城市之间的连通性,则不要发出警报。

输入格式:

输入在第一行给出两个整数N(0 < N <=500)和M(<=5000),分别为城市个数(于是默认城市从0到N-1编号)和连接两城市的通路条数。随后M行,每行给出一条通路所连接的两个城市的编号,其间以1个空格分隔。在城市信息之后给出被攻占的信息,即一个正整数K和随后的K个被攻占的城市的编号。

注意:输入保证给出的被攻占的城市编号都是合法的且无重复,但并不保证给出的通路没有重复。

输出格式:

对每个被攻占的城市,如果它会改变整个国家的连通性,则输出“Red Alert: City k is lost!”,其中k是该城市的编号;否则只输出“City k is lost.”即可。如果该国失去了最后一个城市,则增加一行输出“Game Over.”。

输入样例:
5 4
0 1
1 3
3 0
0 4
5
1 2 0 4 3
输出样例:
City 1 is lost.
City 2 is lost.
Red Alert: City 0 is lost!
City 4 is lost.
City 3 is lost.
Game Over.
分析:用图的深度优先遍历判断一个图内的连通分量有多少个,标记为cnt,之后对于每一个输入数据,因为城市a被攻占,所以把a的所有路径标注为不可达(0),再统计连通分量的个数tempcnt,如果tempcnt > cnt + 1,也就是说当现在的连通分量多余以前的连通分量+1的时候,说明改变了图的连通性;(因为城市被攻占本身它城市自己就变成了一个单独的城市,多出来一个连通分量,只要tempcint <= cnt + 1都说明没有改变图的连通性),每一次tempcnt在用完之后把cnt的值更新为tempcnt,保证下一次的判断是建立再已经失去之前这么多城市的基础之上的。
因为题目中说输入保证给出的被攻占的城市编号都是合法的且无重复,所以如果城市失去了n个,就是当前输入的是从0开始的第n-1个数据的时候,就说明Game Over了,最后当if(i == n – 1) printf(“Game Over.\n”);

 

L2-001. 紧急救援-PAT团体程序设计天梯赛GPLT(Dijkstra算法)

作为一个城市的应急救援队伍的负责人,你有一张特殊的全国地图。在地图上显示有多个分散的城市和一些连接城市的快速道路。每个城市的救援队数量和每一条连接两个城市的快速道路长度都标在地图上。当其他城市有紧急求助电话给你的时候,你的任务是带领你的救援队尽快赶往事发地,同时,一路上召集尽可能多的救援队。

输入格式:

输入第一行给出4个正整数N、M、S、D,其中N(2<=N<=500)是城市的个数,顺便假设城市的编号为0~(N-1);M是快速道路的条数;S是出发地的城市编号;D是目的地的城市编号。第二行给出N个正整数,其中第i个数是第i个城市的救援队的数目,数字间以空格分隔。随后的M行中,每行给出一条快速道路的信息,分别是:城市1、城市2、快速道路的长度,中间用空格分开,数字均为整数且不超过500。输入保证救援可行且最优解唯一。

输出格式:

第一行输出不同的最短路径的条数和能够召集的最多的救援队数量。第二行输出从S到D的路径中经过的城市编号。数字间以空格分隔,输出首尾不能有多余空格。

输入样例:
4 5 0 3
20 30 40 10
0 1 1
1 3 2
0 3 3
0 2 2
2 3 2
输出样例:
2 60
0 1 3
分析:用一遍dijkstra算法。设立num[i]和w[i]表示从出发点到i结点拥有的路的条数,以及能够找到的救援队的数目~~当判定dis[u] + e[u][v] < dis[v]的时候,不仅仅要更新dis[v],还要更新num[v] = num[u], w[v] = weight[v] + w[u]; 如果dis[u] + e[u][v] == dis[v],还要更新num[v] += num[u],而且判断一下是否权重w[v]更小,如果更小了就更新w[v] = weight[v] + w[u];
再设立一个pre[i]表示最短路径的前一个结点,在dis[u] + e[u][v] <= dis[v]的时候更新pre[v] = u,最后递归打印路径即可

 

1003. Emergency (25)-PAT甲级真题(Dijkstra算法)

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) – the number of cities (and the cities are numbered from 0 to N-1), M – the number of roads, C1 and C2 – the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output

2 4

题目大意:n个城市m条路,每个城市有救援小组,所有的边的边权已知。给定起点和终点,求从起点到终点的最短路径条数以及最短路径上的救援小组数目之和。如果有多条就输出点权(城市救援小组数目)最大的那个~

分析:用一遍Dijkstra算法~救援小组个数相当于点权,用Dijkstra求边权最小的最短路径的条数,以及这些最短路径中点权最大的值~dis[i]表示从出发点到i结点最短路径的路径长度,num[i]表示从出发点到i结点最短路径的条数,w[i]表示从出发点到i点救援队的数目之和~当判定dis[u] + e[u][v] < dis[v]的时候,不仅仅要更新dis[v],还要更新num[v] = num[u], w[v] = weight[v] + w[u]; 如果dis[u] + e[u][v] == dis[v],还要更新num[v] += num[u],而且判断一下是否权重w[v]更小,如果更小了就更新w[v] = weight[v] + w[u]; 

1016. Phone Bills (25)-PAT甲级真题

A long-distance telephone company charges its customers by the following rules:

Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a long-distance call, the time will be recorded, and so will be the time when the customer hangs up the phone. Every calendar month, a bill is sent to the customer for each minute called (at a rate determined by the time of day). Your job is to prepare the bills for each month, given a set of phone call records.

Input Specification:

Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.

The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 – 01:00, the toll from 01:00 – 02:00, and so on for each hour in the day.

The next line contains a positive number N (<= 1000), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word “on-line” or “off-line”.

For each test case, all dates will be within a single month. Each “on-line” record is paired with the chronologically next record for the same customer provided it is an “off-line” record. Any “on-line” records that are not paired with an “off-line” record are ignored, as are “off-line” records not paired with an “on-line” record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.

Output Specification:

For each test case, you must print a phone bill for each customer.

Bills must be printed in alphabetical order of customers’ names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.

Sample Input:
10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10
10
CYLL 01:01:06:01 on-line
CYLL 01:28:16:05 off-line
CYJJ 01:01:07:00 off-line
CYLL 01:01:08:03 off-line
CYJJ 01:01:05:59 on-line
aaa 01:01:01:03 on-line
aaa 01:02:00:01 on-line
CYLL 01:28:15:41 on-line
aaa 01:05:02:24 on-line
aaa 01:04:23:59 off-line
Sample Output:
CYJJ 01
01:05:59 01:07:00 61 $12.10
Total amount: $12.10
CYLL 01
01:06:01 01:08:03 122 $24.40
28:15:41 28:16:05 24 $3.85
Total amount: $28.25
aaa 01
02:00:01 04:23:59 4318 $638.80
Total amount: $638.80

分析:将给出的数据先按照姓名排序,再按照时间的先后顺序排列,这样遍历的时候,前后两个名字相同且前面的状态为on-line后面一个的状态为off-line的就是合格数据~

注意:【关于最后一个测试点】计算费用从00:00:00到dd:hh:mm计算可以避免跨天的问题,比如01:12:00到02:02:00