UVA - 414 Machined Surfaces

Machined Surfaces

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

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 [code]XXXX XXXXX

XXX XXXXXXX

XXXXX XXXX

XX XXXXXX

. .

. .

. .

XXXX XXXX

XXX XXXXXX

1 25 [/code]

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. The ``void" 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 (character "B" for ease of reading. The actual input file will use the ASCII-space character, not "B").

4
XXXXBBBBBBBBBBBBBBBBXXXXX
XXXBBBBBBBBBBBBBBBXXXXXXX
XXXXXBBBBBBBBBBBBBBBBXXXX
XXBBBBBBBBBBBBBBBBBXXXXXX
2
XXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXX
1
XXXXXXXXXBBBBBBBBBBBBBBXX
0

Sample
Output

4
0
0

分析：
题目的大致意思是：首先输入一个数字N，代表接下来要输入几行字符串；    每行字符串的格式为：由字母X和空格 （“ ”）组成，且两边是字符X，中间是空格。每行的开始有一个或多个X，中间有0个或多个空格，末端有一个或多个X。    要求：将N行中的左边的字符X同时向右移动（或右边的字符X向左移动），直到有一行中间没有空格时停止所有行的移动，此时计算N行中总共有多少个空格。大体思路：做这道题目的思路就是把每一次输入的字符串的空格数加起来减去最小的一个*输入的行数。。。

代码：// uva414.cpp : 定义控制台应用程序的入口点。
//

//#include "stdafx.h"
#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int main()
{

int k, m, n, i, j, sum, min;
char str[100000];
while (scanf("%d%*c", &n) && n)
{
sum = 0;
for (i = 0; i<n; i++)
{
gets(str);
m = strlen(str);
k = 0;
for (j = 0; j<m; j++)
{
if (str[j] == ' ')
k++;
}
if (i == 0)
min = k;
else
{
if (min>k)
min = k;
}
sum += k;
}
printf("%d\n", sum - min*n);
}
return 0;
}