poj解题报告——poj 1493 Machined Surfaces

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原题入口

poj 1493 Machined Surfaces

题目描述

Machined Surfaces Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 3696 Accepted: 2268

Description An imaging device furnishes digital images of two machined surfaces that eventually will be assembled in contact with each other. The roughness of this final contact is to be estimated.

A digital image is composed of the two characters, “X” and ” ” (space). There are always 25 columns to an image, but the number of rows, N, is variable. Column one (1) will always have an “X” in it and will be part of the left surface. The left surface can extend to the right from column one (1) as contiguous X’s.

Similarly, column 25 will always have an “X” in it and will be part of the right surface. The right surface can extend to the left from column 25 as contiguous X’s.

Digital-Image View of Surfaces

Left Right XXXX XXXXX ←1 XXX XXXXXXX XXXXX XXXX XX XXXXXX . . . . . . XXXX XXXX XXX XXXXX ←N ↑ ↑ 1 25

In each row of the image, there can be zero or more space characters separating the left surface from the right surface. There will never be more than a single blank region in any row.

For each image given, you are to determine the total void" that will exist after the left surface has been brought into contact with the right surface. Thevoid” is the total count of the spaces that remains between the left and right surfaces after theyhave been brought into contact.

The two surfaces are brought into contact by displacing them strictly horizontally towards each other until a rightmost “X” of the left surface of some row is immediately to the left of the leftmost “X” of the right surface of that row. There is no rotation or twisting of these two surfaces as they are brought into contact; they remain rigid, and only move horizontally.

Note: The original image may show the two surfaces already in contact, in which case no displacement enters into the contact roughness estimation.

Input The input consists of a series of digital images. Each image data set has the following format:

First line - A single unsigned integer, N, with value greater than zero (0) and less than 13. The first digit of N will be the first character on a line.

Next N lines - Each line has exactly 25 characters; one or more X’s, then zero or more spaces, then one or more X’s.

The end of data is signaled by a null data set having a zero on the first line of an image data set and no further data.

Output For each image you receive as a data set, you are to reply with the total void (count of spaces remaining after the surfaces are brought into contact). Use the default output for a single integer on a line.

Sample Input

4 XXXX XXXXX XXX XXXXXXX XXXXX XXXX XX XXXXXX 2 XXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX 1 XXXXXXXXX XX 0

Sample Output

4 0 0

Source East Central North America 1995

解题代码

#include <iostream> #include <cstring> using namespace std; #define STRING_LEN 26 char szInputMsg[STRING_LEN] = {0}; int main() { int N; while((cin >> N) && N) { cin.get(); // 去除'\n' int nTotalSapce = 0; int nMinLineSapce = STRING_LEN; for (int nLine = 0; nLine < N; ++nLine) { cin.getline(szInputMsg, STRING_LEN); int nLineSpaceCount = 0; for (int nIndex = 0; nIndex < STRING_LEN - 1; ++nIndex) { if (szInputMsg[nIndex] == ' ') ++nLineSpaceCount; else if (nLineSpaceCount > 0) break; } nTotalSapce += nLineSpaceCount; if (nLineSpaceCount < nMinLineSapce) nMinLineSapce = nLineSpaceCount; } cout << nTotalSapce - (N * nMinLineSapce) << endl; } return 0; }

解题过程

题意:一幅图片是由N行字符串构成,每一行由字母X和空格构成,共有25个字符,并且每行字符以X开头并且以X结尾,由中间连续的空格(0个-23个)分隔,分为左串和右串,题问同时移动各行左串,当有一行的左右X相接时,N行字符串中间的空格有多少!

思路:找出N行的字符串中空格最少的一行,记录其个数”MinSapce”,然后用最初的空格总数减去N*MinSapce即可得到索要的输出结果。

提交结果

AlbertS 认证博客专家 Python C/C 分布式 一个手残党怀着对游戏的好奇心踏上了开发之路,旅途中磕磕绊绊踩了不少坑,触了不少雷,深知好记性不如烂笔头的道理,于是记录下学习路上的点点滴滴,携码起舞,与君共勉。Coding是件有趣的事情,快乐的看待每一天,我真的非常幸福~
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