自动换行问题
给出了一个单词序列，每行的字符数有限制。通过放置换行符，以使打印线清晰可见。
这些行必须平衡，当某些行具有很多额外的空间并且某些行包含少量额外的空间时，它将平衡它们到单独的行。它尝试使用相同数量的额外空间来使它们平衡。
该算法将产生一行中可以放置多少个单词，以及需要多少行。
输入输出
Input:
The length of words for each line. {3, 2, 2, 5}. The max width is 6.
Output:
Line number 1: Word Number: 1 to 1 (only one word)
Line number 2: Word Number: 2 to 3 (Second and 3rd word)
Line number 3: Word Number: 4 to 4 (4th word)算法
wordWrap(wordLenArr, size, maxWidth)
输入- 单词长度数组，数组大小和单词的最大宽度。
输出- 每行将放置多少个单词的列表。
Begin
define two square matrix extraSpace and lineCost of order (size + 1)
define two array totalCost and solution of size (size + 1)
for i := 1 to size, do
extraSpace[i, i] := maxWidth – wordLenArr[i - 1]
for j := i+1 to size, do
extraSpace[i, j] := extraSpace[i, j-1] – wordLenArr[j - 1] - 1
done
done
for i := 1 to size, do
for j := i+1 to size, do
if extraSpace[i, j] < 0, then
lineCost[i, j] = ∞
else if j = size and extraSpace[i, j] >= 0, then
lineCost[i, j] := 0
else
linCost[i, j] := extraSpace[i, j]^2
done
done
totalCost[0] := 0
for j := 1 to size, do
totalCost[j] := ∞
for i := 1 to j, do
if totalCost[i-1] ≠∞ and linCost[i, j] ≠ ∞ and
(totalCost[i-1] + lineCost[i,j] < totalCost[j]), then
totalCost[i – 1] := totalCost[i – 1] + lineCost[i, j]
solution[j] := i
done
done
display the solution matrix
End示例
#include<iostream>
using namespace std;
int dispSolution (int solution[], int size) {
int k;
if (solution[size] == 1)
k = 1;
else
k = dispSolution (solution, solution[size]-1) + 1;
cout << "Line number "<< k << ": Word Number: " <<solution[size]<<" to "<< size << endl;
return k;
}
void wordWrap(int wordLenArr[], int size, int maxWidth) {
int extraSpace[size+1][size+1];
int lineCost[size+1][size+1];
int totalCost[size+1];
int solution[size+1];
for(int i = 1; i<=size; i++) { //find extra space for all lines
extraSpace[i][i] = maxWidth - wordLenArr[i-1];
for(int j = i+1; j<=size; j++) { //extra space when word i to j are in single line
extraSpace[i][j] = extraSpace[i][j-1] - wordLenArr[j-1] - 1;
}
}
for (int i = 1; i <= size; i++) { //find line cost for previously created extra spaces array
for (int j = i; j <= size; j++) {
if (extraSpace[i][j] < 0)
lineCost[i][j] = INT_MAX;
else if (j == size && extraSpace[i][j] >= 0)
lineCost[i][j] = 0;
else
lineCost[i][j] = extraSpace[i][j]*extraSpace[i][j];
}
}
totalCost[0] = 0;
for (int j = 1; j <= size; j++) { //find minimum cost for words
totalCost[j] = INT_MAX;
for (int i = 1; i <= j; i++) {
if (totalCost[i-1] != INT_MAX && lineCost[i][j] != INT_MAX && (totalCost[i-1] + lineCost[i][j] < totalCost[j])){
totalCost[j] = totalCost[i-1] + lineCost[i][j];
solution[j] = i;
}
}
}
dispSolution(solution, size);
}
main() {
int wordLenArr[] = {3, 2, 2, 5};
int n = 4;
int maxWidth = 6;
wordWrap (wordLenArr, n, maxWidth);
}输出结果
Line number 1: Word Number: 1 to 1 Line number 2: Word Number: 2 to 3 Line number 3: Word Number: 4 to 4