UVA 1347(POJ 2677)Tour(双调欧几里得旅行商问题)

最新推荐文章于 2017-05-05 22:54:35 发布

原创最新推荐文章于 2017-05-05 22:54:35 发布 · 894 阅读

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#dp #uva

本文介绍了一种解决双调欧几里得旅行商问题的动态规划方法，该问题要求找到连接一系列按x坐标升序排列的点的最短闭合路径。文章提供了完整的C++实现代码，通过动态规划策略计算从起点出发，先左后右回到起点的最短路径长度。

Tour

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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 p_i = < x_i, y_i > . 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.

Note: 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

Sample Output

6.47
7.89

题意：典型的动态规划例题。又叫做双调欧几里得旅行商问题。算法导论里面的题目。

思路：

dp[i][j] 表示从 i 到 1，再从1到j的距离。在这个路径上，点 1 到 Pmax(i,j) 点之间的所有点有且仅有经过一次。

dp[i][j] = dp[i-1][j] + dis(i,i-1);

dp[i][i-1] = min (dp[i][i-1], dp[i-1][j] + dis(i, j));

<span style="font-size:18px;">
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const int INF = 1<<29;
const int MAXN = 1100;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
struct node
{
    double x;
    double y;
}a[MAXN];
double dp[MAXN][MAXN];

int cmp(node a, node b)
{
    return a.x < b.x;
}

double dist(int i, int j)
{
    return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n;
    while(cin>>n)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%lf %lf", &a[i].x, &a[i].y);
        }
        sort(a+1, a+1+n, cmp);
        dp[2][1] = dist(1, 2);
        for(int i = 3; i <= n; i++)
        {
            dp[i][i-1] = INF*1.0;
            for(int j = 1; j < i-1; j++)
            {
                dp[i][i-1] = min(dp[i][i-1], dp[i-1][j]+dist(i, j));
                dp[i][j] = dp[i-1][j]+dist(i-1, i);
            }
        }
        double ans = INF*1.0;
        for(int i = 1; i < n; i++)
        {
            ans = min(ans, dp[n][i]+dist(n, i));
        }
        printf("%.2f\n", ans);
    }
    return 0;
}
</span>