【鏈接】 我是鏈接,點我呀:)
【題意】

給你k個紅包,每個紅包可以在si..ti的時間範圍內拿走。
搶完紅包之後你得到wi元,然後你需要在di+1時刻才能繼續搶紅包
時間是線性的從1..n
然後某個人可以阻止你在x時刻搶紅包,然後你的時間跳過1s(-1s)直接到達x+1時刻.
這個人可以阻止你m次。
請問這個人采用最優阻止策略下,你最少搶到的金額。
(如果有多個可以搶的紅包,那麽搶wi最大的,如果仍然相同搶di最大的,再相同的話就無所謂了,因為選哪個都一樣了)

【題解】

dp
設dp[i][j]表示i時刻已經用了j次阻止機會,後面能搶到的最少金額.
我們可以用sort+優先隊列求出choose[i]

每次取最小值就好
記得開long

【代碼】

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;
        //打擾t時刻
        long temp1 = (long)1e17;
        if (cnt+1<=m) temp1 = Math.min(temp1, dfs(t+1,cnt+1));
        
        //不打擾
        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;//排完序再標號!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());
        }
    }
}
 

【Codeforces 1106E】Lunar New Year and Red Envelopes