GDUT_寒假训练题解报告_专题II_个人题解报告——题目:J -畅通工程续

TTL电路搭建半加器实验:操作指南完整版 通过TTL电路搭建半加器,详解逻辑设计与实验步骤,帮助掌握数字电路中的半加器工作原理与实际应用,适合电子技术初学者动手实践。 阅读详情

GDUT_寒假训练题解报告_专题II_个人题解报告——题目:J -畅通工程续

题目:

省自从实行了很多年的畅通工程计划后,终于修建了很多路。不过路多了也不好,每次要从一个城镇到另一个城镇时,都有许多种道路方案可以选择,而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。

现在,已知起点和终点,请你计算出要从起点到终点,最短需要行走多少距离。

Input

本题目包含多组数据,请处理到文件结束。
每组数据第一行包含两个正整数N和M(0<N<200,0<M<1000),分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0~N-1编号。
接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。
再接下一行有两个整数S,T(0<=S,T<N),分别代表起点和终点。

Output

对于每组数据,请在一行里输出最短需要行走的距离。如果不存在从S到T的路线,就输出-1.

Sample Input

3 3
0 1 1
0 2 3
1 2 1
0 2
3 1
0 1 1
1 2

Sample Output

2
-1

审完题目,发现是一个板子floyd,直接上手板子,但是要注意一个问题:这**题目会给一些鬼畜数据:同样的两个顶点,给出的距离是更小的,那么输入就是要有这样的一句话:

edge[from][to]=min(edge[from][to],value);

极其狗的数据,然后就是板子floyd没了,数据是点集大小小于200,不要开100
完整代码:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>

#define INF 0x3f3f3f3f
#define ULL unsigned long long
#define LL long long
using namespace std;
int edge[205][205];
int n,m;

int main()
{
	while(~scanf("%d %d",&n,&m))
	{
		memset(edge,INF,sizeof(edge));
		for(int time=0;time<n;time++)
		{
			edge[time][time]=0;;
		}
		for(int time=0;time<m;time++)
		{
			int from,to,value;
			scanf("%d %d %d",&from,&to,&value);
			edge[from][to]=min(edge[from][to],value);
			edge[to][from]=edge[from][to];
		}
		int S,F;
		scanf("%d %d",&S,&F);
		for(int time=0;time<n;time++)
		{
			for(int time1=0;time1<n;time1++)
			{
				for(int time2=0;time2<n;time2++)
				{
					edge[time1][time2]=min(edge[time1][time2],edge[time1][time]+edge[time][time2]);

				}
			}
		}
		/*for(int time=0;time<n;time++)
		{
			for(int time1=0;time1<n;time1++)
			{
				printf("%d ",edge[time][time1]);
			}
		}*/
		if( edge[S][F]==INF )printf("-1\n");
		else printf("%d\n",edge[S][F]);

	}
	return 0;
}

YOLOv8 道路病害检测全栈工程|22 类路面裂缝坑洞 VOC&YOLO 数据集、PyQt 精美 GUI、ONNX 量化、全套训练评估曲线落地 城市道路、国省道、高速公路路面长期受车辆碾压、雨水侵蚀,衍生纵向 / 横向裂纹、坑洼、车辙、接缝剥落等二十余种路面病害。传统人工道路巡检存在致命行业痛点:人工徒步 / 乘车目视巡检效率极低,单日有效巡查里程不足 20 公里;毫米级细微线性裂纹肉眼极易漏检,雨水冲刷后微小病害快速扩张为大坑槽,养护成本翻倍;强光暴晒、阴雨积水、路面油污场景人工分辨难度大;海量巡检图片人工分类统计病害台账耗时数天,养护决策缺乏量化数据支撑。传统阈值、边缘检测图像处理仅能识别大面积坑洞,无法区分二十余种细分裂纹类型,泛化能力极差。 阅读详情

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% Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

GDUT_排位赛题解报告_第3场_H. Photoshoot

题目: Farmer John is lining up his N cows (2≤N≤103), numbered 1…N, for a photoshoot. FJ initially planned for the i-th cow from the left to be the cow numbered ai, and wrote down the permutation a1,a2,…...

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GDUT_寒假训练题解报告_专题I_B、C、D题 个人题解报告

GDUT_寒假训练题解报告_专题I_B、C、D题 个人题解报告 B题 :dfs (我更喜欢叫它感染函数,因为之前写扫雷的时候教别人,不敢说dfs就说感染函数说顺口了。) 由于最近下雨,水在农民约翰的田里的各个地方汇集起来,用一个N×M(1<=N<=100;1<=M<=100)正方形表示。每一个正方形包含水(“W”)或旱地(“.”)。农场主约翰想弄清楚他的田里有多少池塘。...

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GDUT_排位赛题解报告_第5场_H. 完美排列

题目: ChenJr定义一个长度为n的完美排列为:对于一个任意的的i,都有a[i−1]>a[i]<a[i+1] 或者 a[i−1]<a[i]>a[i+1] 给你一个正整数n,你需要输出一个字典序最小的完美排列。 Input 一行输入一个正整数n (1≤n≤105)。 Output 输出n个整数,代表答案的排列。 Example inputCopy 1 outputCopy ...

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GDUT_寒假训练题解报告_数论专题_个人题解报告——题目:D - Beautiful Numbers (CodeForces - 300C)【逆元、组合数公式】

原题链接http://codeforces.com/problemset/problem/300/C 题目: Vitaly is a very weird man. He’s got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal representation of this int...

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GDUT_排位赛题解报告_第3场_D. Race

Bessie is running a race of length K (1≤K≤109) meters. She starts running at a speed of 0 meters per second. In a given second, she can either increase her speed by 1 meter per second, keep it unchang...

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GDUT_寒假训练题解报告_数论专题_个人题解报告——题目:E - Revenge of GCD【HDU - 5019】【数论(GCD,因数分解)】

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=5019 题目: In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), highest common factor (hcf), or greatest com...

DevourPower的博客 432

GDUT_排位赛题解报告_第4场_C. Defining Labels

题目: Microsoft Excel is a spreadsheet developed by Microsoft. It features calculation, graphing tools, pivot tables, and a macro programming language called Visual Basic for Applications. It has been a...

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GDUT_排位赛题解报告_第1场_B. MooBuzz

题目: Farmer John’s cows have recently become fans of playing a simple number game called “FizzBuzz”. The rules of the game are simple: standing in a circle, the cows sequentially count upward from one,...

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GDUT_排位赛题解报告_第3场_B.Loan Repayment

题目: Farmer John owes Bessie N gallons of milk (1≤N≤1012). He has to give her the milk within K days. However, he doesn’t want to give the milk away too quickly. On the other hand, he has to make forwa...

DevourPower的博客 408

GDUT_排位赛题解报告_第5场_A. 唯一排列

题目 ChenJr给你一个长度为n的排列p,你可以交换其中任意相邻的两个数。现在你需要用最小的交换次数使得这个排列变成升序的序列。现在ChenJr想知道,对于这个排列p,是否存在唯一的交换方式,使得这个排列变成升序的序列。 Input 第一行包含一个正整数T (1≤T≤5∗104),代表有T组样例。 对于每组样例,第一行包含一个n (1≤∑n≤105),代表排列的大小。 下一行包含n个正整数pi,...

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