题目链接

这道题我感觉状态定义不太好想，需要一定的经验

import java.util.*;
/**
 * 蜗牛
 * 状态定义：
 * dp[i][0]:到达(x[i],0)最小时间
 * dp[i][1]:到达 xi 上方的传送门最小时间
 */

public class Main {
    static Scanner in = new Scanner(System.in);
    static final int N = 100010,INF = 0x3f3f3f3f;
    static int n;
    static int[] x = new int[N];
    static double[][] dp = new double[N][2];
    static int[] a,b;
    public static void main(String[] args) {
        n = in.nextInt();
        a = new int[n+1];
        b = new int[n+1];
        for (int i = 1;i <= n;i++) {
            x[i] = in.nextInt();
        }
        for (int i = 1;i <= n-1;i++) {
            a[i] = in.nextInt();
            b[i] = in.nextInt();
        }
        for (int i = 2;i <= n;i++) {
            dp[i][0] = dp[i][1] = INF;
        }
        dp[1][0] = x[1];
        dp[1][1] = x[1] + a[1]/0.7;
        for (int i = 2;i <= n;i++) {
            // 到达i底部有两种办法
            // 1.从i-1底部爬过来
            // 2.从i-1的传送门过来，然后再爬下来
            dp[i][0] = Math.min(dp[i-1][0] + x[i] - x[i-1],dp[i-1][1] + b[i-1]/1.3);
        
            // 到达i上方传送门有两种办法
            // 1.从i-1的传送门过来，然后向上or向下爬
            // 2.从i底部爬上来
            double time = 0.0; // 从i-1传送过来的位置爬到i的传送门的时间
            if (b[i-1] < a[i]) time = (a[i] - b[i-1])/0.7;
            else time = (b[i-1] - a[i])/1.3;
            dp[i][1] = Math.min(dp[i][0] + a[i]/0.7,dp[i-1][1] + time);
        }
        System.out.printf("%.2f",dp[n][0]);
    }
}