POJ2137——Cowties

Cowties
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 2870 Accepted: 961

Description

N cows (3 <= N <= 100) are eating grass in the middle of a field. So that they don't get lost, Farmer John wants to tie them together in a loop so that cow i is attached to cows i-1 and i+1. Note that cow N will be tied to cow 1 to complete the loop.

Each cow has a number of grazing spots she likes and will only be happy if she ends up situated at one of these spots. Given that Farmer John must ensure the happiness of his cows when placing them, compute the shortest length of rope he needs to tie them all in a loop. It is possible for different parts of the loop to cross each other.

Input

* Line 1: The integer N.

* Lines 2..N+1: Each line describes one cow using several space-separated integers. The first integer is the number of locations S (1 <= S <= 40) which are preferred by that cow. This is followed by 2*S integers giving the (x,y) coordinates of these locations respectively. The coordinates lie in the range -100..100.

Output

A single line containing a single integer, 100 times the minimum length of rope needed (do not perform special rounding for this result).

Sample Input

4
1 0 0
2 1 0 2 0
3 -1 -1 1 1 2 2
2 0 1 0 2

Sample Output

400

Hint

[Cow 1 is located at (0,0); cow 2 at (1,0); cow 3 at (1,1); and cow 4 at (0,1).]

Source

USACO 2003 February Orange


n头牛,每头牛都有其限制的放牧区域,求给n头牛安排好放牧区域以后,他们之间的最短距离(形成一个环)

很明显的dp,但是由于头尾相接,所以不能单纯的从1枚举到n,一开始我是这么做的,然后wa了2次,晚上睡觉的时候在想这个,然后就想到了枚举1这头牛的可放牧区域,然后从2枚举到n,求出最小值即可,最后注意答案*100之后不要用%.0f输出,直接强制类型转换成int输出就可以了

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  

using namespace std;

double dp[110][50];

vector<pair<int, int> >pp[110];

double dist(int x1, int y1, int x2, int y2)
{
	return sqrt((double)((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)) );
}

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		int cnt, x, y;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &cnt);
			for (int j = 0; j < cnt; ++j)
			{
				scanf("%d%d", &x, &y);
				pp[i].push_back( make_pair(x, y) );
			}
		}
		memset (dp, 0, sizeof(dp) );
		double ans = 0x3f3f3f3f;
		int fc = pp[1].size();
		for (int i = 0; i < fc; ++i) // 枚举第一行
		{
			int sc = pp[2].size();
			for (int p = 0; p < sc; ++p)
			{
				dp[2][p] = dist(pp[1][i].first, pp[1][i].second, pp[2][p].first, pp[2][p].second);
			}
			for (int j = 3; j <= n; ++j)
			{
				int tc = pp[j].size();
				for (int k = 0; k < tc; ++k)
				{
					int ffc = pp[j - 1].size();
					double mins = 0x3f3f3f3f;
					 for (int q = 0; q < ffc; ++q)
					 {
					 	mins = min(mins, dp[j - 1][q] + dist(pp[j - 1][q].first, pp[j - 1][q].second, pp[j][k].first, pp[j][k].second));
					 }
					 dp[j][k] = mins;
					 if (j == n)
					{
						ans = min(ans, dp[n][k] + dist(pp[n][k].first, pp[n][k].second, pp[1][i].first, pp[1][i].second));	
					}
				}
			}
		}
		printf("%d\n", int(ans * 100));
	}
	return 0;
}


资源下载链接为: https://pan.quark.cn/s/1bfadf00ae14 “STC单片机电压测量”是一个以STC系列单片机为基础的电压检测应用案例,它涵盖了硬件电路设计、软件编程以及数据处理等核心知识点。STC单片机凭借其低功耗、高性价比和丰富的I/O接口,在电子工程领域得到了广泛应用。 STC是Specialized Technology Corporation的缩写,该公司的单片机基于8051内核,具备内部振荡器、高速运算能力、ISP(在系统编程)和IAP(在应用编程)功能,非常适合用于各种嵌入式控制系统。 在源代码方面,“浅雪”风格的代码通常简洁易懂,非常适合初学者学习。其中,“main.c”文件是程序的入口,包含了电压测量的核心逻辑;“STARTUP.A51”是启动代码,负责初始化单片机的硬件环境;“电压测量_uvopt.bak”和“电压测量_uvproj.bak”可能是Keil编译器的配置文件备份,用于设置编译选项和项目配置。 对于3S锂电池电压测量,3S锂电池由三节锂离子电池串联而成,标称电压为11.1V。测量时需要考虑电池的串联特性,通过分压电路将高电压转换为单片机可接受的范围,并实时监控,防止过充或过放,以确保电池的安全和寿命。 在电压测量电路设计中,“电压测量.lnp”文件可能包含电路布局信息,而“.hex”文件是编译后的机器码,用于烧录到单片机中。电路中通常会使用ADC(模拟数字转换器)将模拟电压信号转换为数字信号供单片机处理。 在软件编程方面,“StringData.h”文件可能包含程序中使用的字符串常量和数据结构定义。处理电压数据时,可能涉及浮点数运算,需要了解STC单片机对浮点数的支持情况,以及如何高效地存储和显示电压值。 用户界面方面,“电压测量.uvgui.kidd”可能是用户界面的配置文件,用于显示测量结果。在嵌入式系统中,用
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