modify code

master
algorithmzuo 2 years ago
parent d0af212a15
commit 1d60dbe1b9

@ -14,15 +14,26 @@ import java.io.StreamTokenizer;
public class Code03_BiggestBSTTopologyInTree {
public static int MAXN = 200001;
public static int[][] tree = new int[MAXN][3];
public static int[] record = new int[MAXN];
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StreamTokenizer in = new StreamTokenizer(br);
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
while (in.nextToken() != StreamTokenizer.TT_EOF) {
int n = (int) in.nval;
for (int i = 1; i <= n; i++) {
tree[i][0] = 0;
tree[i][1] = 0;
tree[i][2] = 0;
record[i] = 0;
}
in.nextToken();
int h = (int) in.nval;
int[][] tree = new int[n + 1][3];
for (int i = 1; i <= n; i++) {
in.nextToken();
int c = (int) in.nval;
@ -35,7 +46,7 @@ public class Code03_BiggestBSTTopologyInTree {
tree[c][1] = l;
tree[c][2] = r;
}
out.println(maxBSTTopology(h, tree, new int[n + 1]));
out.println(maxBSTTopology(h));
out.flush();
}
}
@ -44,37 +55,26 @@ public class Code03_BiggestBSTTopologyInTree {
// t: 代表树 t[i][0]是i节点的父节点、t[i][1]是i节点的左孩子、t[i][2]是i节点的右孩子
// m: i节点为头的最大bst拓扑结构大小 -> m[i]
// 返回: 以h为头的整棵树上最大bst拓扑结构的大小
public static int maxBSTTopology(int h, int[][] t, int[] m) {
public static int maxBSTTopology(int h) {
if (h == 0) {
return 0;
}
int l = t[h][1];
int r = t[h][2];
int c = 0;
int p1 = maxBSTTopology(l, t, m);
int p2 = maxBSTTopology(r, t, m);
while (l < h && m[l] != 0) {
l = t[l][2];
int l = tree[h][1];
int r = tree[h][2];
int p1 = maxBSTTopology(l);
int p2 = maxBSTTopology(r);
int tmp = l;
while (tmp < h && record[tmp] != 0) {
tmp = tree[tmp][2];
}
if (m[l] != 0) {
c = m[l];
while (l != h) {
m[l] -= c;
l = t[l][0];
}
}
while (r > h && m[r] != 0) {
r = t[r][1];
}
if (m[r] != 0) {
c = m[r];
while (r != h) {
m[r] -= c;
r = t[r][0];
}
record[l] -= record[tmp];
tmp = r;
while (tmp > h && record[tmp] != 0) {
tmp = tree[tmp][1];
}
m[h] = m[t[h][1]] + m[t[h][2]] + 1;
return Math.max(Math.max(p1, p2), m[h]);
record[r] -= record[tmp];
record[h] = record[l] + record[r] + 1;
return Math.max(Math.max(p1, p2), record[h]);
}
}

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