Hdu 2224 The shortest path（双调TSP）

最新推荐文章于 2020-04-24 20:06:48 发布

Spark_007

最新推荐文章于 2020-04-24 20:06:48 发布

阅读量1.3k

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划 2014寒假训练文章标签：双调TSP 动态规划

本文链接：https://blog.youkuaiyun.com/spark_007/article/details/19493757

动态规划同时被 2 个专栏收录

24 篇文章

订阅专栏

2014寒假训练

9 篇文章

订阅专栏

解决HDOJ平台上的2224号问题，即求解给定点集上的最短路径，需遵循特定的访问规则。通过动态规划算法计算最优路径长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2224

The shortest path

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 654 Accepted Submission(s): 335

Problem Description

There are n points on the plane, Pi(xi, yi)(1 <= i <= n), and xi < xj (i<j). You begin at P1 and visit all points then back to P1. But there is a constraint:
Before you reach the rightmost point Pn, you can only visit the points those have the bigger x-coordinate value. For example, you are at Pi now, then you can only visit Pj(j > i). When you reach Pn, the rule is changed, from now on you can only visit the points those have the smaller x-coordinate value than the point you are in now, for example, you are at Pi now, then you can only visit Pj(j < i). And in the end you back to P1 and the tour is over.
You should visit all points in this tour and you can visit every point only once.

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n(2 <= n <= 200), means the number of points. Then following n lines each containing two positive integers Pi(xi, yi), indicating the coordinate of the i-th point in the plane.

Output

For each test case, output one line containing the shortest path to visit all the points with the rule mentioned above.The answer should accurate up to 2 decimal places.

Sample Input

Sample Output

  
   6.47

Hint: The way 1 - 3 - 2 - 1 makes the shortest path.

题意：给n个二维点，你要先沿x坐标严格递增的方向从1点走到n点，然后在沿x坐标严格递减的方向从n点走到1点，求走过的路程和的最小值。

分析：可以将上述的一条路看成从1点到n点的没有交叉点的两条路A和B。两条路分别从点1 向x递增的方向 A走到 i，B走到 j 。

就可以定义 dp[i][j] 为A走到 i ，B走到 j 的最短距离。题目就变成了求dp[n][n]的值。

转移方程：

dp[1][2] = D[i][j]; //i ,j 两点的真实距离。

i < j-1 时； dp[ i ][ j ] = dp[ i ][ j-1] + D[ j-1 ][ j ];

i == j-1 时；dp[ i ][ j ] = min(dp[ k ][ j-1 ] + D[ k ][ j ]);

dp[ n ][ n ] = dp[ n-1 ][ n ]+ D[ n-1 ][ n ].

代码：

#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAX 205
#define inf 9999999
double dp[MAX][MAX];
double D[MAX][MAX];
struct node
{
	double x,y;
}p[MAX];
int cmp(const void *a,const void *b)
{
	struct node *p = (struct node *)a;
	struct node *q = (struct node *)b;
	return (p->x - q->x >0)?1:-1;
}
double dis(node a,node b)
{
	return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
}
void init(int n)
{
	for (int i = 1; i <= n; i++)
	for (int j = 1; j <= n; j++)
	{
		D[i][j] = inf;
	}
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		init(n);
		for (int i = 1;i<=n;i++)
		{
			scanf("%lf %lf",&p[i].x,&p[i].y);
			for(int j=1;j<i;j++)
				D[i][j]=D[j][i] = dis(p[i],p[j]);//记录直接距离。
		}
		qsort(p+1,n,sizeof(p[1]),cmp);
		dp[1][2] = D[1][2];
		for (int i = 3; i <= n; i++)
		{
			for (int j = 1; j <= i-2; j++)//when i-1 in the higher dir
			{
				dp[j][i] = dp[j][i-1] + D[i-1][i];
			}
			double Min = inf;
			for (int j = 1; j <= i-2; j++)//when i-1 in the lower dir
			{
				 Min = min(dp[j][i-1]+D[j][i],Min);
			}dp[i-1][i] = Min;
		}
		dp[n][n] = dp[n-1][n]+D[n][n-1];
		printf("%.2lf\n",dp[n][n]);
	}
	return 0;
}