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java - GSS1 -SPOJ - 线段树 TLE

转载 作者:塔克拉玛干 更新时间:2023-11-03 06:27:08 27 4
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我正在解决 this problem通过使用线段树。我在每个节点保存总和、最大值、最左边的最大值和最右边的最大值。然后我搜索图表以找到特定时间间隔的答案。我怎样才能提高这段代码的速度?

import java.util.Scanner;
//TLE
class GSS1 {


static class Node{
int max;
int MaxL;
int MaxR;
int sum;

public Node(int max, int MaxL, int MaxR, int sum){
this.max=max;
this.MaxL=MaxL;
this.MaxR=MaxR;
this.sum=sum;
}

public Node(){

}
}

static class SegmentTree{

private Node[] tree;
private int maxsize;
private int height;

private final int STARTINDEX = 0;
private final int ENDINDEX;
private final int ROOT = 0;
Node s;

public SegmentTree(int size){
height = (int)(Math.ceil(Math.log(size) / Math.log(2)));
maxsize = 2 * (int) Math.pow(2, height) - 1;
tree = new Node[maxsize];
for(int i=0;i<tree.length;i++){
tree[i]=new Node();
}
ENDINDEX = size - 1;
s=new Node();
s.MaxL=Integer.MIN_VALUE;
s.MaxR=Integer.MIN_VALUE;
s.sum=Integer.MIN_VALUE;
s.max=Integer.MIN_VALUE;

}




private int leftchild(int pos){
return 2 * pos + 1;
}

private int rightchild(int pos){
return 2 * pos + 2;
}

private int mid(int start, int end){
return (start + (end - start) / 2);
}

private Node constructSegmentTreeUtil(int[] elements, int startIndex, int endIndex, int current){
if (startIndex == endIndex)
{
tree[current].max=tree[current].MaxL=tree[current].MaxR=tree[current].sum=elements[startIndex];
return tree[current];
}
int mid = mid(startIndex, endIndex);
Node left=constructSegmentTreeUtil(elements, startIndex, mid, leftchild(current));
Node right=constructSegmentTreeUtil(elements, mid + 1, endIndex, rightchild(current));
tree[current].max = Math.max(left.max, right.max);
tree[current].MaxL = Math.max(left.MaxL , left.sum+right.MaxL);
tree[current].MaxR = Math.max(right.MaxR , right.sum+left.MaxR);
tree[current].sum = left.sum+right.sum;
return tree[current];
}

public void constructSegmentTree(int[] elements){
constructSegmentTreeUtil(elements, STARTINDEX, ENDINDEX, ROOT);
}

private Node getSumUtil(int startIndex, int endIndex, int queryStart, int queryEnd, int current){

if (queryStart <= startIndex && queryEnd >= endIndex ){
return tree[current];
}
if (endIndex < queryStart || startIndex > queryEnd){
return s;
}
int mid = mid(startIndex, endIndex);

Node left=getSumUtil(startIndex, mid, queryStart, queryEnd, leftchild(current));
Node right=getSumUtil( mid + 1, endIndex, queryStart, queryEnd, rightchild(current));

Node current_Node=new Node();
current_Node.max = Math.max(left.max, right.max);
current_Node.MaxL = Math.max(left.MaxL , left.sum+right.MaxL);
current_Node.MaxR = Math.max(right.MaxR , right.sum+left.MaxR);
current_Node.sum = left.sum+right.sum;
return current_Node;


}

public int getMaxSum(int queryStart, int queryEnd){
if(queryStart < 0 || queryEnd > tree.length)
{System.out.println("inside negative");
return Integer.MIN_VALUE;
}
return getMax(getSumUtil(STARTINDEX, ENDINDEX, queryStart, queryEnd, ROOT));
}

public int getMax(Node r){
return Math.max(Math.max(r.max, r.MaxL),Math.max(r.MaxR, r.sum));
}

public int getFirst(){
return tree[0].MaxL;
}

}


public static void main(String[] args) {
Scanner input=new Scanner(System.in);

int numbers[]=new int [input.nextInt()];

for(int i=0;i<numbers.length;i++){
numbers[i]=input.nextInt();
}

SegmentTree tree=new SegmentTree(numbers.length);
tree.constructSegmentTree(numbers);

int cases=input.nextInt();

int x;
int y;
int query;
for(int i=0;i<cases;i++){
x=input.nextInt()-1;
y=input.nextInt()-1;

System.out.println(tree.getMaxSum(x, y));
}






}

}

最佳答案

您的方法是正确的,但 I/O 速度对于这个问题也很重要,因为时间限制非常严格。您应该使用自定义阅读器,因为 Scanner 非常慢。使用下面提到的类来读取输入。

class Reader {
final private int BUFFER_SIZE = 1 << 16;private DataInputStream din;private byte[] buffer;private int bufferPointer, bytesRead;
public Reader(){din=new DataInputStream(System.in);buffer=new byte[BUFFER_SIZE];bufferPointer=bytesRead=0;
}public Reader(String file_name) throws IOException{din=new DataInputStream(new FileInputStream(file_name));buffer=new byte[BUFFER_SIZE];bufferPointer=bytesRead=0;
}public String readLine() throws IOException{byte[] buf=new byte[64];int cnt=0,c;while((c=read())!=-1){if(c=='\n')break;buf[cnt++]=(byte)c;}return new String(buf,0,cnt);
}public int nextInt() throws IOException{int ret=0;byte c=read();while(c<=' ')c=read();boolean neg=(c=='-');if(neg)c=read();do{ret=ret*10+c-'0';}while((c=read())>='0'&&c<='9');if(neg)return -ret;return ret;
}public long nextLong() throws IOException{long ret=0;byte c=read();while(c<=' ')c=read();boolean neg=(c=='-');if(neg)c=read();do{ret=ret*10+c-'0';}while((c=read())>='0'&&c<='9');if(neg)return -ret;return ret;
}public double nextDouble() throws IOException{double ret=0,div=1;byte c=read();while(c<=' ')c=read();boolean neg=(c=='-');if(neg)c = read();do {ret=ret*10+c-'0';}while((c=read())>='0'&&c<='9');if(c=='.')while((c=read())>='0'&&c<='9')ret+=(c-'0')/(div*=10);if(neg)return -ret;return ret;
}private void fillBuffer() throws IOException{bytesRead=din.read(buffer,bufferPointer=0,BUFFER_SIZE);if(bytesRead==-1)buffer[0]=-1;
}private byte read() throws IOException{if(bufferPointer==bytesRead)fillBuffer();return buffer[bufferPointer++];
}public void close() throws IOException{if(din==null) return;din.close();}
}

关于java - GSS1 -SPOJ - 线段树 TLE,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23377354/

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