您的位置:首页 > 其它

POJ 3667 线段树 区间合并

2016-10-13 10:51 288 查看


2 a b:将[a,a+b-1]的房间清空


1 a:询问是不是有连续长度为a的空房间,有的话住进最左边

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class Main {

public static void main(String[] args) {
new POJ3667().solve() ;
}
}

class POJ3667{
InputReader in = new InputReader(System.in) ;
PrintWriter out = new PrintWriter(System.out) ;
SegTree tree = new SegTree() ;

void solve(){
int n = in.nextInt() ;
int m = in.nextInt() ;
tree.build(1, 1, n) ;
while(m-- > 0){
if(in.nextInt() == 1){
int v = in.nextInt() ;
if(tree.sum[1] < v) out.println(0) ;
else{
int d = tree.query(v, 1, 1, n) ;
out.println(d) ;
tree.update(d, d+v-1, 0, 1, 1, n) ;
}
}
else{
int d = in.nextInt() ;
int w = in.nextInt() ;
tree.update(d , d+w-1, 1, 1, 1, n) ;
}
}
out.flush() ;
}

}

class SegTree{
int N = 50008 ;
int[] sum = new int[N<<2] ;
int[] lsum = new int[N<<2] ;
int[] rsum = new int[N<<2] ;
int[] color = new int[N<<2] ;
int[] len = new int[N<<2] ;

void down(int t){
if(color[t] != -1){
color[t<<1] = color[t<<1|1] = color[t] ;
sum[t<<1] = lsum[t<<1] = rsum[t<<1] = color[t] * len[t<<1] ;
sum[t<<1|1] = lsum[t<<1|1] = rsum[t<<1|1] = color[t] * len[t<<1|1] ;
color[t] = -1 ;
}
}

void up(int t){
lsum[t] = (lsum[t<<1] == len[t<<1]) ? len[t<<1] + lsum[t<<1|1] : lsum[t<<1]  ;
rsum[t] = (rsum[t<<1|1] == len[t<<1|1]) ? len[t<<1|1] + rsum[t<<1] : rsum[t<<1|1] ;
sum[t] = Math.max(sum[t<<1] , sum[t<<1|1]) ;
sum[t] = Math.max(sum[t], rsum[t<<1] + lsum[t<<1|1]) ;
}

void build(int t , int l , int r){
color[t] = -1 ;
sum[t] = lsum[t] = rsum[t] = len[t] = r - l + 1 ;
if(len[t] == 1) return ;
int m = (l + r) >> 1 ;
build(t<<1 , l , m) ;
build(t<<1|1, m+1, r) ;
}

void update(int L , int R , int V , int t , int l , int r){
if(L <= l && r <= R){
lsum[t] = rsum[t] = sum[t] = len[t] * V  ;
color[t] = V ;
return ;
}
down(t) ;
int m = (l + r) >> 1 ;
if(L <= m) update(L, R, V, t<<1, l, m) ;
if(R > m) update(L, R, V, t<<1|1, m+1 , r) ;
up(t) ;
}

int query(int V , int t , int l , int r){
if(l == r) return l ;
down(t) ;
int m = (l + r) >> 1 ;
if(sum[t<<1] >= V) return query(V, t<<1 , l, m) ;
else if(rsum[t<<1] + lsum[t<<1|1] >= V) return m - rsum[t<<1] + 1 ;
else return query(V, t<<1|1, m+1 , r) ;
}
}

class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;

public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = new StringTokenizer("");
}

private void eat(String s) {
tokenizer = new StringTokenizer(s);
}

public String nextLine() {
try {
return reader.readLine();
} catch (Exception e) {
return null;
}
}

public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String s = nextLine();
if (s == null)
return false;
eat(s);
}
return true;
}

public String next() {
hasNext();
return tokenizer.nextToken();
}

public int nextInt() {
return Integer.parseInt(next());
}

public long nextLong() {
return Long.parseLong(next());
}

public double nextDouble() {
return Double.parseDouble(next());
}

public BigInteger nextBigInteger() {
return new BigInteger(next());
}

}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: