HDU 2224 The shortest path

最新推荐文章于 2021-03-08 07:30:06 发布

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最新推荐文章于 2021-03-08 07:30:06 发布

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原文链接：http://www.cnblogs.com/zlrrrr/p/9873122.html

本文介绍了一种解决平面点集上特定约束条件下的最短路径问题的算法。问题设定为从左到右访问所有点后返回起点，且访问顺序受限。通过动态规划和距离矩阵，算法求解了满足规则的最短路径，并提供了代码实现。

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

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

1 1

2 3

3 1

Sample Output

6.47

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

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
double dist[210][210];
double dp[210][210];

struct Node {
    int x;
    int y;
}node[210];

double dis(int a, int b) {
    return sqrt((double)(node[a].x - node[b].x) * (node[a].x - node[b].x) + (node[a].y - node[b].y) * (node[a].y - node[b].y));
}

int main() {
    while(~scanf("%d", &N)) {
        for(int i = 1; i <= N; i ++) {
            scanf("%d%d", &node[i].x, &node[i].y);
            for(int j = 1; j < i; j ++)
                dist[i][j] = dist[j][i] = dis(i, j);
        }

        memset(dp, 99999999, sizeof(dp));

        dp[1][2] = dp[2][1] = dist[1][2];
        for(int j = 3; j <= N; j ++) {
            dp[1][j] = dp[j][1] = dp[1][j - 1] + dist[j - 1][j];
        }

        for(int i = 2; i <= N; i ++) {
            double ans = 99999999;
            for(int k = 1; k < i; k ++)
                ans = min(ans, dp[i][k] + dist[k][i + 1]);

            dp[i][i + 1] = dp[i + 1][i] = ans;
            for(int j = i + 2; j <= N; j ++)
                dp[i][j] = dp[j][i] = dp[i][j - 1] + dist[j - 1][j];
        }
        dp[N][N] = dp[N][N - 1] + dist[N - 1][N];
        printf("%.2lf\n", dp[N][N]);
    }

    return 0;
}

转载于:https://www.cnblogs.com/zlrrrr/p/9873122.html