HDU----1155

Description

Once again, James Bond is fleeing from some evil people who want to see him dead. Fortunately, he has left a bungee rope on a nearby highway bridge which he can use to escape from his enemies. His plan is to attach one end of the rope to the bridge, the other end of the rope to his body and jump off the bridge. At the moment he reaches the ground, he will cut the rope, jump into his car and be gone. 

Unfortunately, he had not had enough time to calculate whether the bungee rope has the right length, so it is not clear at all what is going to happen when he jumps off the bridge. There are three possible scenarios: 
The rope is too short (or too strong), and James Bond will never reach the ground. 
The rope is too long (or too weak), and James Bond will be going too fast when he touches the ground. Even for a special agent, this can be very dangerous. You may assume that if he collides at a speed of more than 10 m/s, he will not survive the impact. 
The rope's length and strength are good. James Bond touches the ground at a comfortable speed and can escape. 
As his employer, you would like to know whether James Bond survives or whether you should place a job ad for the soon-to-be vacant position in the local newspaper. Your physicists claim that: 
The force with which James is pulled towards the earth is 
9.81 * w, 
where w is his weight in kilograms and 9.81 is the Earth acceleration in meters over squared seconds. 
Mr. Bond falls freely until the rope tautens. Then the force with which the bungee rope pulls him back into the sky depends on the current length of the rope and is 
k * Δl, 
where Δl is the difference between the rope's current length and its nominal, unexpanded length, and k is a rope-specific constant. 
Given the rope's strength k, the nominal length of the rope l in meters, the height of the bridge s in meters, and James Bond's body weight w, you have to determine what is going to happen to our hero. For all your calculations, you may assume that James Bond is a point at the end of the rope and the rope has no mass. You may further assume that k, l, s, and w are non-negative and that s < 200. 

The input contains several test cases, one test case per line. Each test case consists of four floating-point numbers (k, l, s, and w) that describe the situation. Depending on what is going to happen, your program must print "Stuck in the air.", "Killed by the impact.", or "James Bond survives.". Input is terminated by a line containing four 0s, this line should not be processed.

Input

350 20 30 75
375 20 30 75
400 20 30 75
425 20 30 75
450 20 30 75
400 20 30 50
400 20 30 80
400 20 30 85
0 0 0 0

Output

Killed by the impact.
James Bond survives. 
James Bond survives. 
James Bond survives. 
Stuck in the air.
Stuck in the air.
James Bond survives. 

Killed by the impact.

这个这就是一个简单题,主要就是物理的能量守恒,输入四个数k, l, s, w分别表示弹簧系数,弹簧长度,形变量, 体重,然后求出重力做功, 然后减去弹簧做功,然后求出速度,判断速度大小

#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<math.h> using namespace std; #define N 1100 #define g 9.81 int main() {     double k, l, s, w;     while(scanf("%lf%lf%lf%lf", &k,&l,&s,&w)!=EOF)     {         if(k==0 && l==0 && s==0 && w==0)             break;         double e=w*g*s;         if(s>l)             e-=k*(s-l)*(s-l)/2;         if(e<0)         {             printf("Stuck in the air.\n");             continue;         }                      else         {             double v;             v=sqrt(2*e/w);             if(v>=10)                 printf("Killed by the impact.\n");             else                 printf("James Bond survives.\n");         }     }     return 0; }

HDU-3480 是一个典型的动态规划问题,其题目标题通常为 *Division*,主要涉及二维费用背包问题或优化后的动态规划策略。题目大意是:给定一个整数数组,将其划分为若干个连续的子集,每个子集最多包含 $ m $ 个元素,并且每个子集的最大值与最小值之差不能超过给定的阈值 $ t $,目标是使所有子集的划分代价总和最小。每个子集的代价是该子集最大值与最小值的差值。 ### 动态规划思路 设 $ dp[i] $ 表示前 $ i $ 个元素的最小代价。状态转移方程如下: $$ dp[i] = \min_{j=0}^{i-1} \left( dp[j] + cost(j+1, i) \right) $$ 其中 $ cost(j+1, i) $ 表示从第 $ j+1 $ 到第 $ i $ 个元素构成一个子集的代价,即 $ \max(a[j+1..i]) - \min(a[j+1..i]) $。 为了高效计算 $ cost(j+1, i) $,可以使用滑动窗口或单调队列等数据结构来维护区间最大值与最小值,从而将时间复杂度优化到可接受的范围。 ### 示例代码 以下是一个简化版本的动态规划实现,使用暴力方式计算区间代价,适用于理解问题结构: ```cpp #include <bits/stdc++.h> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 10010; int a[MAXN]; int dp[MAXN]; int main() { int T, n, m; cin >> T; for (int Case = 1; Case <= T; ++Case) { cin >> n >> m; for (int i = 1; i <= n; ++i) cin >> a[i]; dp[0] = 0; for (int i = 1; i <= n; ++i) { dp[i] = INF; int mn = a[i], mx = a[i]; for (int j = i; j >= max(1, i - m + 1); --j) { mn = min(mn, a[j]); mx = max(mx, a[j]); if (mx - mn <= T) { dp[i] = min(dp[i], dp[j - 1] + mx - mn); } } } cout << "Case " << Case << ": " << dp[n] << endl; } return 0; } ``` ### 优化策略 - **单调队列**:可以使用两个单调队列分别维护当前窗口的最大值与最小值,从而将区间代价计算的时间复杂度从 $ O(n^2) $ 降低到 $ O(n) $。 - **斜率优化**:若问题满足特定的决策单调性,可以考虑使用斜率优化技巧进一步加速状态转移过程。 ### 时间复杂度分析 原始暴力解法的时间复杂度为 $ O(n^2) $,在 $ n \leq 10^4 $ 的情况下可能勉强通过。通过单调队列优化后,可以稳定运行于 $ O(n) $ 或 $ O(n \log n) $。 ### 应用场景 HDU-3480 的问题模型可以应用于资源调度、任务划分等场景,尤其适用于需要控制子集内部差异的问题,如图像分块压缩、数据分段处理等[^1]。 ---
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