Codeforces Round 932 (Div. 2) --- C. Messenger in MAC --- 题解

C Messenger in MAC

题目大意:

思路解析:

答案计算为 , 可以发现当所选的几个信息固定后,其实后面的一项就变为b_max - b_min,得到了这个结论之后,其实我们可以直接把整个信息按照b进行排序,枚举l,r,那么我最多能选的信息的限制就变为了 a的和 <= L - (br -bl),

因为我们需要选择最多的信息,所以我们在l-r中选择尽量a较小的信息。但是我们可以把选择较小,转为当我们在还能选择时就把当前的选择,当超过时,就不选择当前已经选择中最大的信息。

这部分代码:(这里set可以使用任意排序的数据结构) 每次选择的东西大于限制时,就弹出已经选择了的最大信息

代码实现:

复制代码
import java.io.*;
import java.util.*;

import static java.lang.String.*;


public class Main {
    static int MAXN = 100005;
    static int n;
    static int mod = 1000000;
    static long INF = (long) 1e18;


    public static void main(String[] args) throws IOException {
        FastScanner f = new FastScanner();
        PrintWriter w = new PrintWriter(System.out);
        int t = f.nextInt();
        for (int o = 0; o < t; o++) {
            int n = f.nextInt();
            int l = f.nextInt();
            int[][]  a = new int[n][2];
            for (int i = 0; i < n; i++) {
                a[i][0] = f.nextInt();
                a[i][1] = f.nextInt();
            }
            Arrays.sort(a, ((o1, o2) -> {
                return o1[1] - o2[1];
            }));
            int max = 0;

            for (int i = 0; i < n; i++) {
                PriorityQueue<Integer> set = new PriorityQueue<>(new Comparator<Integer>() {
                    @Override
                    public int compare(Integer o1, Integer o2) {
                        return o2 - o1;
                    }
                });
                long cur = 0;

                for (int j = i; j < n; j++) {
                    if (a[j][1] - a[i][1] > l) break;
                    cur += a[j][0];
                    set.add(a[j][0]);
                    while(a[j][1] - a[i][1] + cur > l){
                        int num = set.poll();
                        cur -= num;
                    }
                    max = Math.max(max, set.size());
                }

            }
            w.println(max);
        }
        w.flush();
        w.close();
    }

    private static class FastScanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner() throws IOException {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return (short) -ret;
            return ret;
        }

        private 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;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ') c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        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++];
        }
    }


}
相关推荐
Wuliwuliii几秒前
贡献延迟计算DP
数据结构·c++·算法·动态规划·dp
苦藤新鸡3 分钟前
2.字母异位词分组
c语言·c++·力扣·哈希算法
ysn111114 分钟前
简单多边形三角剖分---耳切法(含源码)
算法
e疗AI产品之路5 分钟前
一文介绍Philips DXL心电图算法
算法·pan-tompkins·心电分析
CryptoRzz9 分钟前
印度交易所 BSE 与 NSE 实时数据 API 接入指南
java·c语言·python·区块链·php·maven·symfony
小袁顶风作案15 分钟前
leetcode力扣——135.分发糖果
算法·leetcode·职场和发展
橘颂TA26 分钟前
【Linux】从 “抢资源” 到 “优雅控场”:Linux 互斥锁的原理与 C++ RAII 封装实战(Ⅰ)
linux·运维·服务器·c++·算法
YGGP41 分钟前
【Golang】LeetCode 19. 删除链表的倒数第 N 个节点
算法·leetcode·链表
枫叶丹441 分钟前
【Qt开发】Qt系统(三)->事件过滤器
java·c语言·开发语言·数据库·c++·qt