bibin
/
cphb
1
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity

Merge 2d53e3a6f5 into f269ae3919

This commit is contained in:
Hung Dang 2019-10-14 15:20:18 +00:00 committed by GitHub
parent f269ae3919 2d53e3a6f5
commit ed6bef7224
No known key found for this signature in database
GPG Key ID: 4AEE18F83AFDEB23
1 changed files with 15 additions and 5 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

20
chapter07.tex
View File

@ -247,7 +247,7 @@ value[0] = 0;
for (int x = 1; x <= n; x++) {
value[x] = INF;
for (auto c : coins) {
if (x-c >= 0) {
if (x >= c) {
value[x] = min(value[x], value[x-c]+1);
}
}
@ -283,7 +283,7 @@ value[0] = 0;
for (int x = 1; x <= n; x++) {
value[x] = INF;
for (auto c : coins) {
if (x-c >= 0 && value[x-c]+1 < value[x]) {
if ((x >= c) && value[x-c]+1 < value[x]) {
value[x] = value[x-c]+1;
first[x] = c;
}
@ -358,7 +358,7 @@ for $0 \le x \le n$:
count[0] = 1;
for (int x = 1; x <= n; x++) {
for (auto c : coins) {
if (x-c >= 0) {
if (x >= c) {
count[x] += count[x-c];
}
}
@ -634,12 +634,22 @@ int sum[N][N];
\end{lstlisting}
and calculate the sums as follows:
\begin{lstlisting}
for (int y = 1; y <= n; y++) {
for (int x = 1; x <= n; x++) {
for (int y = 1; y < N; y++) {
for (int x = 1; x < N; x++) {
sum[y][x] = max(sum[y][x-1],sum[y-1][x])+value[y][x];
}
}
\end{lstlisting}
However, below is a better solution that only uses one dimensional
array to keep track of the maximum sums
\begin{lstlisting}
for (int y = 1; y < N; ++y) {
sum[0] += value[y][0];
for (int x = 1; x < N; ++x) {
sum[x] = value[y][x] + max(sum[x-1], sum[x]);
}
}
\end{lstlisting}
The time complexity of the algorithm is $O(n^2)$.
\section{Knapsack problems}