1 条题解

  • 1
    @ 2025-9-21 10:05:30
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    
    const int maxn=500000;
    const long long neInf=0x8080808080808080;
    struct gezi {
        int juli;
        int zhi;
    } a[maxn+1];
    long long dp[maxn+1];
    int q[maxn+1];
    int n,d,k,lbound,rbound,ans=-1;
    long long sum;
    
    void kuaidu(int &p) {
        char c;
        int f=1;
        p=0;
        do {
            c=getchar();
            if (c=='-')
                f=-1;
        } while (c<'0'||c>'9');
        do p=p*10+c-'0', c=getchar();
        while (c>='0'&&c<='9');
        p=p*f;
    }
    
    void init() {
        cin>>n>>d>>k;
        for (int i=1; i<=n; i++) {
            kuaidu(a[i].juli);
            kuaidu(a[i].zhi);
            if (a[i].zhi>0)
                sum+=a[i].zhi;
        }
        rbound=max(a[n].juli,d);
    }
    
    long long dynamic_programming(int zuo, int you) {
        memset(dp,0x80,sizeof(dp));
        dp[0]=0;
        memset(q,0,sizeof(q));
        int tou=1, wei=0, j=0;
        /*for (int i=1; i<=n; i++)
            for (int j=0; j<i; j++)
                if (a[i].juli-a[j].juli>=zuo&&a[i].juli-a[j].juli<=you&&dp[j]!=neInf)
                    dp[i]=max(dp[i],dp[j]+a[i].zhi);*/
        for (int i=1; i<=n; i++) {
            while (a[i].juli-a[j].juli>=zuo&&j<i) {
                if (dp[j]!=neInf) {
                    while (tou<=wei&&dp[q[wei]]<=dp[j])
                        wei--;
                    q[++wei]=j;
                }
                j++;
            }
            while (tou<=wei&&a[i].juli-a[q[tou]].juli>you)
                tou++;
            if (tou<=wei)
                dp[i]=dp[q[tou]]+a[i].zhi;
        }
        long long num=neInf;
        for (int i=1; i<=n; i++)
            if (dp[i]>num)
                num=dp[i];
        return num;
    }
    
    int main() {
        //freopen("jump.in","r",stdin);
        //freopen("jump.out","w",stdout);
        init();
        if (sum<k) {
            cout<<"-1"<<endl;
            return 0;
        }
        while (lbound<=rbound) {
            int mid=(lbound+rbound)/2;
            int zuobianjie=max(1,d-mid);
            int youbianjie=d+mid;
            long long num=dynamic_programming(zuobianjie,youbianjie);
            if (num<k)
                lbound=mid+1;
            else {
                ans=mid;
                rbound=mid-1;
            }
        }
        cout<<ans<<endl;
        //fclose (stdin);
        //fclose (stdout);
        return 0;
    }
    
    • 1

    信息

    ID
    768
    时间
    1000ms
    内存
    256MiB
    难度
    8
    标签
    递交数
    33
    已通过
    7
    上传者