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Hung Dang 2019-10-14 15:20:18 +00:00 committed by GitHub
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chapter07.tex
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@ -247,7 +247,7 @@ value[0] = 0;
for (int x = 1; x <= n; x++) { for (int x = 1; x <= n; x++) {
value[x] = INF; value[x] = INF;
for (auto c : coins) { for (auto c : coins) {
if (x-c >= 0) { if (x >= c) {
value[x] = min(value[x], value[x-c]+1); value[x] = min(value[x], value[x-c]+1);
} }
} }
@ -283,7 +283,7 @@ value[0] = 0;
for (int x = 1; x <= n; x++) { for (int x = 1; x <= n; x++) {
value[x] = INF; value[x] = INF;
for (auto c : coins) { 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; value[x] = value[x-c]+1;
first[x] = c; first[x] = c;
} }
@ -358,7 +358,7 @@ for $0 \le x \le n$:
count[0] = 1; count[0] = 1;
for (int x = 1; x <= n; x++) { for (int x = 1; x <= n; x++) {
for (auto c : coins) { for (auto c : coins) {
if (x-c >= 0) { if (x >= c) {
count[x] += count[x-c]; count[x] += count[x-c];
} }
} }
@ -634,12 +634,22 @@ int sum[N][N];
\end{lstlisting} \end{lstlisting}
and calculate the sums as follows: and calculate the sums as follows:
\begin{lstlisting} \begin{lstlisting}
for (int y = 1; y <= n; y++) { for (int y = 1; y < N; y++) {
for (int x = 1; x <= n; x++) { for (int x = 1; x < N; x++) {
sum[y][x] = max(sum[y][x-1],sum[y-1][x])+value[y][x]; sum[y][x] = max(sum[y][x-1],sum[y-1][x])+value[y][x];
} }
} }
\end{lstlisting} \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 $O(n^2)$. The time complexity of the algorithm is $O(n^2)$.
\section{Knapsack problems} \section{Knapsack problems}