POJ--3158

Kickdown
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 2305 Accepted: 984
Description

A research laboratory of a world-leading automobile company has received an order to create a special transmission mechanism, which allows for incredibly efficient kickdown — an operation of switching to lower gear. After several months of research engineers found that the most efficient solution requires special gears with teeth and cavities placed non-uniformly. They calculated the optimal flanks of the gears. Now they want to perform some experiments to prove their findings.

The first phase of the experiment is done with planar toothed sections, not round-shaped gears. A section of length n consists of n units. The unit is either a cavity of height h or a tooth of height 2h. Two sections are required for the experiment: one to emulate master gear (with teeth at the bottom) and one for the driven gear (with teeth at the top).

There is a long stripe of width 3h in the laboratory and its length is enough for cutting two engaged sections together. The sections are irregular but they may still be put together if shifted along each other.

The stripe is made of an expensive alloy, so the engineers want to use as little of it as possible. You need to find the minimal length of the stripe which is enough for cutting both sections simultaneously.

Input

There are two lines in the input file, each contains a string to describe a section. The first line describes master section (teeth at the bottom) and the second line describes driven section (teeth at the top). Each character in a string represents one section unit — 1 for a cavity and 2 for a tooth. The sections can not be flipped or rotated.

Each string is non-empty and its length does not exceed 100.

Output

Write a single integer number to the output file — the minimal length of the stripe required to cut off given sections.

Sample Input

sample input #1
2112112112
2212112

sample input #2
12121212
21212121

sample input #3
2211221122
21212
Sample Output

sample output #1
10

sample output #2
8

sample output #3
15

大致的意思就是给你两串由‘1’‘2’构成的字符串(字符数组每一个字符都代表了该位置上材料的高度 然后呢工人要用一块每个位置高度都是‘3’的铁皮去装我们所给的两个字符串 让我们输出这个铁皮的最小长度

这是一道水题 刚开始是在CB上面遇到的 先是在CB上面提交了无数次都是WA 去POJ上面提交就AC 奇怪 找错误也没找到 废了一下午

#include<set>
#include<map>
#include<string.h>
#include<string>
#include<vector>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define pi 3.1415926535
#define e 2.718281828459
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f

using namespace std;

const int maxn=220;

char s[maxn],t[maxn];
int compare(char s[],char t[])
{
    int i,j,k;
    int len1=strlen(s);
    int len2=strlen(t);
    for(k=0; k<len1; k++)
    {
        for(i=k,j=0; j<len2; i++,j++)
        {
            if(s[i]+t[j]>=100)
                break;
        }
        if(j==len2)
        {
            int ans=max(i,len1);
            return ans;
            break;
        }
    }
    return len1+len2;
}
int main()
{
    while(scanf("%s%s",s,t)!=EOF)
    {
        int ans=min(compare(s,t),compare(t,s));
        cout<<ans<<endl;
    }
    return 0;
}
/*
2112112112
2212112
12121212
21212121
2211221122
21212
2211221122
2112112
2211221122
211
2211221122
2122112
*/
### POJ 2488 - A Knight’s Journey 的 Python 解决方案 对于骑士之旅问题,在棋盘上找到一条路径使得骑士能够访问每一个位置恰好一次。这个问题可以通过深度优先搜索 (DFS) 来解决。 #### 使用 DFS 实现骑士之旅算法 为了实现这一目标,可以定义一个递归函数来尝试从当前位置出发的所有可能移动方向,并标记已经访问过的位置防止重复访问[^2]。 ```python def dfs(x, y, movei): global board, n, m if all(len(steps) == m * n for steps in board): return True for i in range(8): next_x = x + row_dir[i] next_y = y + col_dir[i] if is_safe(next_x, next_y, board): board[next_x][next_y] = chr(ord('a') + movei) if dfs(next_x, next_y, movei + 1): return True # Backtrack board[next_x][next_y] = '-' return False ``` 此代码片段展示了如何通过回溯法寻找解决方案。`is_safe()` 函数用于验证下一步是否合法;而 `row_dir[]` 和 `col_dir[]` 数组则存储了八个潜在的方向向量供骑士跳跃使用[^5]。 #### 初始化与边界条件处理 在调用上述递归之前,需要初始化一些变量并设置好起始状态: - 创建大小为 N×M 的二维列表作为棋盘表示。 - 设置初始坐标 `(startX,startY)` 并将其设为起点字符 `'A'` 或其他指定标志位。 - 定义终止条件:当所有格子都被遍历到时返回成功结果。 #### 输出格式化 一旦找到了有效路径,则按照给定样例中的方式打印输出。如果无法完成整个旅程,则报告失败情况如 “impossible”。 ```python if solveKTUtil(board, startX, startY, move_count=0): print_solution() else: print("Scenario #: impossible") ```
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