hdu 3631 Shortest Path(Floyd)[通俗易懂]

hdu 3631 Shortest Path(Floyd)

大家好,又见面了,我是全栈君。

题目链接:Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3962    Accepted Submission(s): 938

Problem Description
When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem.

There is a weighted directed multigraph G. And there are following two operations for the weighted directed multigraph:

(1) Mark a vertex in the graph.

(2) Find the shortest-path between two vertices only through marked vertices.

For it was the first time that LMY faced such a problem, she was very nervous. At this moment, YY decided to help LMY to analyze the shortest-path problem. With the help of YY, LMY solved the problem at once, admiring YY very much. Since then, when LMY meets problems, she always calls YY to analyze the problems for her. Of course, YY is very glad to help LMY. Finally, it is known to us all, YY and LMY become programming lovers.

Could you also solve the shortest-path problem?

 
Input
The input consists of multiple test cases. For each test case, the first line contains three integers N, M and Q, where N is the number of vertices in the given graph, N≤300; M is the number of arcs, M≤100000; and Q is the number of operations, Q ≤100000. All vertices are number as 0, 1, 2, … , N – 1, respectively. Initially all vertices are unmarked. Each of the next M lines describes an arc by three integers (x, y, c): initial vertex (x), terminal vertex (y), and the weight of the arc (c). (c > 0) Then each of the next Q lines describes an operation, where operation “0 x” represents that vertex x is marked, and operation “1 x y” finds the length of shortest-path between x and y only through marked vertices. There is a blank line between two consecutive test cases.

End of input is indicated by a line containing N = M = Q = 0.

 
Output
Start each test case with “Case #:” on a single line, where # is the case number starting from 1.

For operation “0 x”, if vertex x has been marked, output “ERROR! At point x”.

For operation “1 x y”, if vertex x or vertex y isn’t marked, output “ERROR! At path x to y”; if y isn’t reachable from x through marked vertices, output “No such path”; otherwise output the length of the shortest-path. The format is showed as sample output.

There is a blank line between two consecutive test cases.

 
Sample Input
   
   
5 10 10 1 2 6335 0 4 5725 3 3 6963 4 0 8146 1 2 9962 1 0 1943 2 1 2392 4 2 154 2 2 7422 1 3 9896 0 1 0 3 0 2 0 4 0 4 0 1 1 3 3 1 1 1 0 3 0 4 0 0 0

 
Sample Output
   
   
Case 1: ERROR! At point 4 ERROR! At point 1 0 0 ERROR! At point 3 ERROR! At point 4

 
Source


代码例如以下:

#include <cstdio>
#include <cstring>
#define INF 0x3fffffff  
#define MAXN 317
int dis[MAXN][MAXN];
int mark[MAXN];
int n;
int min(int a, int b)
{
	return a < b ? a:b;
}
void init()
{
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(i == j)
				dis[i][j] = 0;
			else
				dis[i][j] = INF;
		}
	}
}

void Floyd(int k)
{
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
		}
	}
}

int main()
{
	int i, j;
	int M, Q;
	int a, b, w;
	int cas = 0;
	while(~scanf("%d%d%d",&n,&M,&Q))
	{
		if(n == 0 && M == 0 && Q == 0)
			break;
		if(cas != 0)
			printf("\n");
		init();
		memset(mark,0,sizeof(mark));
		for(i =  0; i < M; i++)
		{
			scanf("%d%d%d",&a,&b,&w);
			if(w < dis[a][b])
			{
				dis[a][b] = w;
			}
		}
		int op, x, y;
		printf("Case %d:\n",++cas);
		for(i = 0; i < Q; i++)
		{
			scanf("%d",&op);
			if(op == 0)
			{
				scanf("%d",&x);
				if(mark[x])
				{
					printf("ERROR! At point %d\n",x);
					continue;
				}
				mark[x] = 1;
				Floyd(x);
			}
			else if(op == 1)
			{
				scanf("%d%d",&x,&y);
				if(!mark[x] || !mark[y])
				{
					printf("ERROR! At path %d to %d\n",x,y);
					continue;
				}
				if(dis[x][y] >= INF)
					printf("No such path\n");
				else
					printf("%d\n",dis[x][y]);
			}
		}
	}
	return 0;
}
版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请联系我们举报,一经查实,本站将立刻删除。

发布者:全栈程序员-站长,转载请注明出处:https://javaforall.net/115664.html原文链接:https://javaforall.net

(0)
全栈程序员-站长的头像全栈程序员-站长


相关推荐

  • mysql的sql分页查询语句怎么写_sql 分页查询语句(mysql分页语句)「建议收藏」

    sql分页查询语句(mysql分页语句)2020-07-2411:18:53共10个回答intpageCount=15(每页显示的行数)intTotalCount=30(页数*每页显示的行数),这里是第二页stringsearchString=xxxxxx(搜索条件)selecttoppageCount*from表名whereidnotin(selecttopTotalCountidfrom表…

    2022年4月13日
    180
  • 视频地址获取

    软件具体名字就不说了哈,首先贴出视频的网页地址:http://www.365yg.com/a6542366077089743367/首先需要获取视频的videoId,直接请求该url,然后match匹配:Patternpattern=Pattern.compile(&amp;quot;videoId:\'(.+)\’&amp;quot;);Matchermatcher=pattern.ma…

    2022年4月8日
    61
  • python浮雕图片_python图片处理PIL

    python浮雕图片_python图片处理PIL一、PIL介绍PIL中所涉及的基本概念有如下几个:通道(bands)、模式(mode)、尺寸(size)、坐标系统(coordinatesystem)、调色板(palette)、信息(info)和滤波器(filters)1、通道每张图片都是由一个或者多个数据通道构成。PIL允许在单张图片中合成相同维数和深度的多个通道。以RGB图像为例,每张图片都是由三个数据通道构成,分别为R、G和B通道。而对…

    2022年6月20日
    29
  • matlab逐行读取字符串txt_matlab批量读取文件并处理

    matlab逐行读取字符串txt_matlab批量读取文件并处理转载自:http://blog.sciencenet.cn/blog-762216-1086021.html%Theloadfunctioncanbeusedtoloadtxtfile,inwhicheachrowhasthesamenumberofelements.%Thisscript(read_line)istoreadthetxtf…

    2025年8月27日
    8
  • python3.7安装pycrypto

    python3.7安装pycrypto首先直接打开 cmd 注意不是打开 python 也不用切换到 python 命令行直接输入 pip3installp 我这里安装过了 所以提示我已经安装过 如果未安装则提示安装成功然后找到你的 python 包的安装目录 也就是上面我们提示的 c users 杨 appdata local programs python python37 lib sit

    2025年6月16日
    5
  • kali装电脑_Kali安装教程(Windows7和kali双系统安装教程)[通俗易懂]

    kali装电脑_Kali安装教程(Windows7和kali双系统安装教程)[通俗易懂]Kali安装教程(一)Windows7+kali双系统安装准备先安装win7,然后安装kali,并且kali的引导项安装在/boot分区而不安装在MBR中,因为这样的话就算重装win7,kali系统经过简单引导仍然可以使用。Win7的安装此处就不在赘述,现在只聊一聊kali的安装。注:教程中个别截图借用了网上某人热心童鞋的截图,在此表示感谢!!1)准备阶段准备4G的优盘一个;刻录工具为:unetb…

    2022年5月5日
    113

发表回复

您的邮箱地址不会被公开。 必填项已用 * 标注

关注全栈程序员社区公众号