【Codeforces 1106E】Lunar New Year and Red Envelopes

给你k个红包,每个红包可以在si..ti的时间范围内拿走。抢完红包之后你得到wi元,然后你需要在di+1时刻才能继续抢红包时间是线性的从1..n 然后某个人可以阻止你在x时刻抢红包,然后你的时间跳过1s(-1s)直接到达x+1时刻. 这个人可以阻止你m次。请问这个人采用最优阻止策略下,你最少抢到的金额。 (如果有多个可以抢的红包,那么抢wi最大的,如果仍然相同抢di最大的,再相同的话就无所谓了,因为选哪个都一样了)

【题解】

dp 设dp[i][j]表示i时刻已经用了j次阻止机会,后面能抢到的最少金额. 我们可以用sort+优先队列求出choose[i] 即表示i时刻你会选择哪一个红包(注意是排序后的红包QAQ) 那么在i时刻有两种选择 1.这个人进行干扰那么转移到dp[i+1][j+1] 2.这个人不进行干扰那么如果i时刻有的选就转移到dp[a[choose[i]].d+1][j]+a[choose[i]].w 如果没得选就转移到dp[i+1][j]

每次取最小值就好
记得开long

【代码】

import java.io.*;
import java.util.*;

public class Main {
	
	static InputReader in;
	static PrintWriter out;
	
	static class RedEnvelope{
		int s,t,d,w,id;
	}
	
	public static Comparator<RedEnvelope> cmp1 = new Comparator<Main.RedEnvelope>() {
		@Override
		public int compare(RedEnvelope o1, RedEnvelope o2) {
			return o1.s-o2.s;
		}
	};
	
	public static Comparator<RedEnvelope> cmp2 = new Comparator<Main.RedEnvelope>() {
		@Override
		public int compare(RedEnvelope o1, RedEnvelope o2) {
			if (o1.w==o2.w) {
				return o2.d-o1.d;
			}else {
				return o2.w-o1.w;
			}
		}
	};
	
	static int N = (int)1e5;
	static int n,m,k;
	static RedEnvelope a[];
	static int choose[];
	static PriorityQueue<RedEnvelope> pq;
	static long dp[][];
	
	static long dfs(int t,int cnt) {
		if (dp[t][cnt]!=-1) return dp[t][cnt];
		if (t>n) return 0;
		//打扰t时刻
		long temp1 = (long)1e17;
		if (cnt+1<=m) temp1 = Math.min(temp1, dfs(t+1,cnt+1));
		
		//不打扰
		long temp2 = (long)1e17;
		if (choose[t]!=-1) {
			temp2 = Math.min(temp2,dfs(a[choose[t]].d+1,cnt)+a[choose[t]].w);
		}else {
			temp2 = Math.min(temp2, dfs(t+1,cnt));
		}
		
		dp[t][cnt] = Math.min(temp1, temp2);
		return dp[t][cnt];
	}
		
	public static void main(String[] args) throws IOException{
		in = new InputReader();
		out = new PrintWriter(System.out);
		
		
		//code start from here
		a = new RedEnvelope[N+10];
		for (int i = 1;i <= N;i++) a[i] = new RedEnvelope();
		
		pq = new PriorityQueue<>(cmp2);
		choose = new int[N+10];
		dp = new long[N+10][200+10];
		for (int i = 0;i <= N;i++)
			for (int j = 0;j <= 200;j++)
					dp[i][j] = -1;
		
		
		n = in.nextInt();m = in.nextInt();k = in.nextInt();

		for (int i = 1;i <= k;i++) {
			a[i].s = in.nextInt();
			a[i].t = in.nextInt();
			a[i].d = in.nextInt();
			a[i].w = in.nextInt();
		}
		Arrays.sort(a, 1,k+1,cmp1);
		for (int i = 1;i <= k;i++) a[i].id = i;//排完序再标号!QAQ
		int j = 1;
		
		for (int i = 1;i <= n;i++) {
			for (;j<=k;) {
				if (a[j].s<=i) {
					pq.add(a[j]);
					j++;
				}else break;
			}
			while (!pq.isEmpty()) {
				RedEnvelope temp = pq.peek();
				if (temp.t<i) {
					pq.poll();
					continue;
				}
				choose[i] = temp.id;
				break;
			}
			if (pq.isEmpty()) choose[i] = -1;
		}
		
		out.println(dfs(1,0));
		out.close();
	}

	static class InputReader{
		public BufferedReader br;
		public StringTokenizer tokenizer;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			tokenizer = null;
		}
		
		public String next(){
			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}