学步园

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.

* 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.

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

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

400

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

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;
}