UVA1347-Tour（动态规划基础）

Problem UVA1347-Tour

Accept: 667  Submit: 3866
Time Limit: 3000 mSec

Problem 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.
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

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

 

Sample Output

6.47

7.89

题解：这个题代码虽然比较短，但是状态的定义真的是很秀，不知道下一次再遇到类似的题自己能不能想到这样定义状态。首先，从左到右再从右到左一般都是化为两个从左到右，dp[i][j]为第一个人到i，第二个人到j，并且标号<=min(i,j)都已经走过，很显然dp[i][j] == dp[j][i]，因此不妨规定i > j，因此dp[i][j]只能从dp[i+1][i]，或dp[i+1][j]转移过来，记忆化搜索就好。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1000 + 10;

int n;

struct Point {
    int x, y;
}point[maxn];

double dist[maxn][maxn];
double dp[maxn][maxn];

double DP(int i, int j) {
    double& ans = dp[i][j];
    if (ans > 0) return ans;

    ans = 0.0;
    if (i == n - 1) {
        ans = dist[n - 1][n] + dist[j][n];
    }
    else {
        ans = min(dist[i][i + 1] + DP(i + 1, j), dist[j][i + 1] + DP(i + 1, i));
    }
    return ans;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    while (~scanf("%d", &n) && n) {
        for (int i = 1; i <= n; i++) {
            scanf("%d%d", &point[i].x, &point[i].y);
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = -1.0;
                dist[j][i] = dist[i][j] = sqrt(1.0*(point[i].x - point[j].x)*(point[i].x - point[j].x) + 1.0*(point[i].y - point[j].y)*(point[i].y - point[j].y));
            }
        }

        printf("%.2f
", DP(1, 1));
    }
    return 0;
}