天啊
2015-11-07 15:15
309 查看
PAT(甲级)在线真题2015/11/15
1003. Emergency (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered(分散的) 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(另一方面), 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(分别地). 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
分析样例:
5 6 0 2
5个城市,6条路,抢救(0+1)1号城市到(2+1)3号之间路;
1 2 1 5 3
1号城市有 1个队,2号2个,3号1个,4号5个,5号3个。
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1
相连的路 路长
输出 2 (1城到3城间最短路有2个) 4 (最短路中最大救援队数是4)
所以这是一道单源最短路径题。用C++比较方便。
待续
1003. Emergency (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered(分散的) 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(另一方面), 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(分别地). 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
分析样例:
5 6 0 2
5个城市,6条路,抢救(0+1)1号城市到(2+1)3号之间路;
1 2 1 5 3
1号城市有 1个队,2号2个,3号1个,4号5个,5号3个。
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1
相连的路 路长
输出 2 (1城到3城间最短路有2个) 4 (最短路中最大救援队数是4)
所以这是一道单源最短路径题。用C++比较方便。
[code]#include<cstido> #include<iostream> #include<queue> #include<algorithm>
待续
相关文章推荐
- 三大框架整合步骤(详细)
- 并发编程之Operation Queue
- 写自己的内核模块——获取一个进程的物理地址
- UIWebView网页视图
- 用JavaScript判断横屏竖屏问题。JavaScript代码如下【转】
- Test-02
- android中LayoutParams的简单理解
- 一个十进制的正整数在二进制表示中的数字个数(C实现)
- LZW压缩算法——简明原理与实现
- 糗事百科爬虫改进
- c 库文件
- Go语言学习笔记-常量
- [置顶] 集群增量会话管理器——DeltaManager
- HDU 4509 hash
- 简单的递归应用
- 内部类
- OFFSET & FETCH
- Cookie / Session / URL重写
- Codeforces 558C Amr and Chemistry
- 泰秘丽人会,女性最放心的丰胸品牌