POJ2677 Tour(DP+双调欧几里得旅行商问题)

Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3929		Accepted: 1761

Description

John Doe, a skilled pilot, enjoys traveling. While on vacation, he rents a small plane and starts visiting beautiful places. To save money, John must determine the shortest closed tour that connects his destinations. Each destination is represented by a point in the plane pi = < xi,yi >. John uses the following strategy: he starts from the leftmost point, then he goes strictly left to right to the rightmost point, and then he goes strictly right back to the starting point. It is known that the points have distinct x-coordinates.
Write a program that, given a set of n points in the plane, computes the shortest closed tour that connects the points according to John's strategy.

Input

The program input is from a text file. Each data set in the file stands for a particular set of points. For each set of points the data set contains the number of points, and the point coordinates in ascending order of the x coordinate. White spaces can occur freely in input. The input data are correct.

Output

For each set of data, your program should print the result to the standard output from the beginning of a line. The tour length, a floating-point number with two fractional digits, represents the result. An input/output sample is in the table below. Here there are two data sets. The first one contains 3 points specified by their x and y coordinates. The second point, for example, has the x coordinate 2, and the y coordinate 3. The result for each data set is the tour length, (6.47 for the first data set in the given example).

Sample Input

3
1 1
2 3
3 1
4
1 1
2 3
3 1
4 2

Sample Output

6.47
7.89

参考博客：http://www.cnblogs.com/-sunshine/archive/2012/07/23/2605251.html

思路【转】：

  欧几里得旅行商问题是对平面上给定的n个点确定一条连接各点的最短闭合旅程的问题。如图（a）给出了一个7个点问题的解。这个问题的一般形式是NP完全的，故其解需要多于多项式的时间。

J.L. Bentley 建议通过只考虑双调旅程(bitonic tour)来简化问题,这种旅程即为从最左点开始，严格地从左到右直至最右点，然后严格地从右到左直至出发点。下图(b)显示了同样的7个点的最短双调路线。在这种情况下，多项式的算法是可能的。事实上，存在确定的最优双调路线的O(n*n)时间的算法。

 图a            图b

注:在一个单位栅格上显示的平面上的七个点。 a)最短闭合路线,长度大约是24.89。这个路线不是双调的。b)相同点的集合上的最短双调闭合路线。长度大约是25.58。

这是一个算导上的思考题15-1。

首先将给出的点排序，关键字x，重新编号，从左至右1，2，3，…，n。

定义p[i][j]，表示结点i到结点j之间的距离。

定义d[i][j]，表示从i连到1，再从1连到j，（注意，i>j，且并没有相连。）

对于任意一个点i来说，有两种连接方法，一种是如图（a）所示，i与i-1相连，另一种呢是如图（b），i与i-1不相连。

根据双调旅程，我们知道结点n一定与n相连，那么，如果我们求的d[n][n-1]，只需将其加上p[n-1][n]就是最短双调闭合路线。

根据上图，很容易写出方程式：

dp[i][j]=dp[i-1][j]+dist[i][i-1];    i这一点是i+1走到的

dp[i][i-1]=min(dp[i][i-1],dp[i-1][j]+dist[j][i]);   i这一点是从j走到的，找到j的最小值

自己的理解：刘汝佳书上的思路看懂了，但是老卡，抽空在想想他的思路，这个思路也好理解，在dp[i][j]这个位置，到下一个i+1，有两种情况要么是i走到的，要么是j走到的，如果是i走到的就对应第一个转移方程，如果是j走到的 dp[i][j] == dp[j][i]，所以对应第二个方程

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAX = 10000 + 10;
struct points
{
    double x,y;
};
points point[MAX];
double d[MAX][MAX];
int cmp(points a, points b)
{
    return (b.x - a.x > 0.00001);
}
double Min(double a, double b)
{
    if(a - b > 0.00001)
        return b;
    else
        return a;
}
double dist(int a, int b)
{
    return sqrt( (point[a].x - point[b].x) * (point[a].x - point[b].x) + (point[a].y - point[b].y) * (point[a].y - point[b].y));
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i++)
            scanf("%lf%lf", &point[i].x, &point[i].y);
        sort(point + 1, point + n + 1, cmp);
        if(n == 1)
        {
            printf("0\n");
            continue;
        }
        d[1][1] = 0;
        for(int i = 1; i <= n; i++)
            d[i][1] = dist(i, 1);

        for(int i = 2; i < n; i++)
        {
            d[i + 1][i] = 10000000.0;
            for(int j = 1; j < i; j++)
            {
                d[i + 1][j] = d[i][j] + dist(i, i + 1);
                d[i + 1][i] = Min(d[i + 1][i], d[i][j] + dist(j,i + 1));
            }
        }
        printf("%.2lf\n", d[n][n - 1] + dist(n - 1, n));
    }
    return 0;
}