zoukankan      html  css  js  c++  java
  • HDU 2224 The shortest path

    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
    3
    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
    ", dp[N][N]);
        }
    
        return 0;
    }
    

      

  • 相关阅读:
    Springboot websocket学习Demo
    webpack与vue使用
    图片服务器图片剪切处理
    时间字段设置默认值
    函数的递归
    数据类型检测及封装
    隔行变色
    if-else案例–开关灯
    作用域
    数据类型核心操作步骤和原理
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9873122.html
Copyright © 2011-2022 走看看