CSU1565-Word Cloud

Word Cloud

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Description

A word cloud (or tag cloud) is a visual representation of textual data based on a weighted metric. In the above cloud (which is based on this year's list of Mid-Central teams), the font size of each word is based
on its number of occurrences in the data set. Tagg Johnson is a man obsessed with counting words that appear in online documents. On his computer, he keeps a spreadsheet of all the sites he visits, along with a list of words that appear on each site and the
number of times such word appears. Tagg would like to generate word clouds based on the data he has collected.

Before describing the algorithm Tagg uses for generating clouds, we digress for a quick lesson in typography. The basic unit of measure is known as a point (typically abbreviated as pt). A font's size is described based on the vertical number of points from
one line to the next, including any interline spacing. For example, with a 12pt font, the vertical space from the top of one character to the top of a character below it is 12 points. We assume that a character's height is precisely equal to the font's point
size (regardless of whether the character is upper or lower case).
For this problem, we focus on a fixed-width font, such as Courier, in which each character of the alphabet is also given the same amount of width. The character width for such a font depends on the font size and
the aspect ratio. For Courier, a word with t characters rendered in a font of size P has a total width of 

 when
measured in points. Note well the use of the ceiling operator, which converts any noninteger to the next highest integer. For example, a 5-letter word in a 20pt font would be rendered with a height of 20 points and a width equal to 

points.
Now we can describe Tagg's algorithm for creating a word cloud. He pre-sorts his word list into alphabetical order and removes words that do not occur at least five times. For each word w, he computes a point size
based on the formula 

, where cw is the number of occurrences of the word, and cmax is the number of occurrences
of the most frequent word in the data set. Note that by this formula, every word will be rendered with anywhere from a 9pt font to a 48pt font. He then places the words in rows, with a 10pt horizontal space between adjacent words, placing as many words as
fit in the row, subject to a maximum width W for his entire cloud. The height of a given row is equal to the maximum font size of any word rendered in that row.
As a tangible example, consider the following data set and word cloud.
Word                   Count
apple          10

banana          5

grape          20

kiwi           18

orange         12

strawberry     10



 


In this example, apple is rendered with 23pt font using width 65pt, banana is rendered with 11pt font using width 38pt, and grape is rendered with 48pt font and width 135pt. If the overall word cloud is constrained
to have width at most 260, those three words fit in a row and the overall height of that row is 48pt (due to grape). On the second row kiwi is rendered with height 43pt and width 97pt, and orange is rendered with height 28pt and width 95pt. A third row has
strawberry with height 23pt and width 130pt. The overall height of this word cloud is 114pt.

Input

 Each data set begins with a line containing two integers: W and N. The value W denotes the maximum width of the cloud; W ≤ 5000 will be at least as wide as any word at its desired font size. The value 1 ≤ N ≤ 100
denotes the number of words that appear in the cloud. Following the first line are N additional lines, each having a string S that is the word (with no whitespace), and an integer C that is a count of the number of occurrences of that word in the original
data set, with 5 ≤ C ≤ 1000. Words will be given in the same order that they are to be displayed within the cloud.

Output

 For each data set, output the word CLOUD followed by a space, a serial number indicating the data set, a colon, another space, and the integer height of the cloud, measured in font points.

Sample Input

260 6
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
250 6
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
610 6
apple 10
banana 5
grape 20
kiwi 18
orange 12
strawberry 10
0 0

Sample Output

CLOUD 1: 114
CLOUD 2: 99
CLOUD 3: 48

Hint

Source

题意：给出n个字符串，每个字符串有一个值，用这个值和题目里给出的方法算出每个字符串的长度和高度，将这些字符串按顺序一个个放入一个长度已知的白纸中，相邻字符串需要间隔10，换行放时需要从上一行的最高高度开始放，问白纸需要多高

解题思路：水题，模拟即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int W, n;
char ch[105][1005];
int h[105], w[105], len[105], a[105];

int main()
{
int cas = 0;
while (~scanf("%d%d", &W, &n) && (n + W))
{
printf("CLOUD %d: ", ++cas);
int ma = 0;
for (int i = 1; i <= n; i++)
{
scanf("%s%d", ch[i], &a[i]);
len[i] = strlen(ch[i]);
ma = max(ma, a[i]);
}
for (int i = 1; i <= n; i++)
{
h[i] = 8 + (int)ceil(1.0 * 40 * (a[i] - 4) / (ma - 4));
w[i] = (int)ceil(1.0 * 9 * len[i] * h[i] / 16);
}
ma = 0;
int k = 0, hh = 0;
for (int i = 1; i <= n; i++)
{
if (k + w[i] <= W) k += 10 + w[i], ma = max(hh + h[i], ma);
else k = w[i] + 10, hh = ma, ma = max(ma, hh + h[i]);
}
printf("%d\n", ma);
}
return 0;
}