Saturday, August 28, 2010


Problem Links:

poj1493, uva00414,


Machined Surfaces
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 1235
Accepted: 837
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. 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.
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.
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
XXXX                XXXXX
XXX               XXXXXXX
XXXXX                XXXX
XX                 XXXXXX
XXXXXXXXX              XX
Sample Output
East Central North America 1995


Just Simulation;

Source Code:

//Tue Mar  9 02:04:56 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int main(int argc, const char* argv[])
//  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
    int n;
    while (cin >> n && n != 0)
        string str;
        int count = 0;
        int localMin = -1;
        for (int i = 0; i < n; i++)
            int cnt = 0;
            getline(cin, str);
            for (int idx = 0; idx < (int) str.size(); idx++)
                cnt += (str[idx] == ' ' ? 1 : 0);
            count += cnt;
            localMin = (localMin > cnt || localMin == -1 ? cnt : localMin);
        cout << count - n * localMin << endl;
//  fclose(stdin);
//  fclose(stdout);
    return 0;

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