add BiggestRectangle

Maximum rectangle under an histogram / in a matrix

code/BiggestRectangle.cpp

@@ -0,0 +1,55 @@
+// INPUT : a boolean matrix
+// OUPUT : the maximum true rectangle (bottom right corner, width, height)
+// COMPLEXITY : linear
+#define MAXX 1000
+// last x, width, height
+tuple<int, int, int> maxUnderHistogram(int sizes[], int N) {
+  stack<pair<int, int> > s; // height, start
+  int mx, mw = 0, mh = 0;
+  for(int i = 0; i < N; i++) {
+    int start = i;
+    while(true) {
+      if(s.empty() || sizes[i] > s.top().first) {
+        s.emplace(sizes[i], start);
+        break;
+      }
+      else if(!s.empty() && sizes[i] < s.top().first) {
+        int w = i - s.top().second;
+        if(w*s.top().first > mw*mh) {
+          mx = i-1; mw = w; mh = s.top().first;
+        }
+        start = s.top().second;
+        s.pop();
+      }
+      else
+        break;
+    }
+  }
+  while(!s.empty()) {
+    int h, st;
+    tie(h, st) = s.top();
+    s.pop();
+    int w = N-st;
+    if(w*h > mw*mh) {
+      mx = N-1; mw = w; mh = h;
+    }
+  }
+  return make_tuple(mx, mw, mh);
+}
+// last x, last y, width, height
+tuple<int, int, int, int> biggestRectangle(bool table[][MAXX], int sizeY, int sizeX) {
+  int mx, my, mw = 0, mh = 0;
+  int mx2, mh2, mw2;
+  int s[MAXX];
+  memset(s, 0, sizeof(s));
+  for(int y = 0; y < sizeY; y++) {
+    for(int x = 0; x < sizeX; x++) {
+      s[x] = table[y][x] ? s[x]+1 : 0;
+    }
+    tie(mx2, mw2, mh2) = maxUnderHistogram(s, sizeX);
+    if(mw2*mh2 > mw*mh) {
+      mx = mx2; my = y; mw = mw2; mh = mh2;
+    }
+  }
+  return make_tuple(mx, my, mw, mh);
+}

main.tex

@@ -131,6 +131,9 @@ Temps de cuisson : $O(n)$
 \subsection{Longest increasing subsequence}
 {\scriptsize\lstinputlisting{code/LongestIncreasingSubsequence.cc}}
 
+\subsection{Maximum rectangle under an histogram / in a matrix}
+{\scriptsize\lstinputlisting{code/BiggestRectangle.cpp}}
+
 \subsection{Dates}
 {\scriptsize\lstinputlisting{code/Dates.cc}}