2 条题解

  • 0
    @ 2025-10-8 16:56:15
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    typedef pair<LL, int> PLI;
    const int N=1e5+100;
    LL a[N];
    int L[N], R[N], del[N];
    priority_queue<PLI, vector<PLI>, greater<PLI>> q;
    void remove(int x)
    {
        L[R[x]]=L[x];
        R[L[x]]=R[x];
        del[x]=1;
    }
    int main()
    {
        int n, m;scanf("%d%d", &n, &m);
        int k=0;
        for(int i=1;i<=n;i++)
    	{
            LL x;scanf("%lld", &x);
            if(x!=0)
    		{
                if(!k || a[k]*x < 0)a[++k]=x;
                else a[k]+=x;
            }
        }
        n=k;
        LL ans=0, cnt=0;
        for(int i=1;i<=n;i++)
    	{
            L[i] = i-1; R[i] = i+1;
            q.push({abs(a[i]), i});
            if(a[i] > 0)ans += a[i], cnt++;
        }
        memset(del, 0, sizeof(del));
        while(cnt > m)
    	{
            while(del[q.top().second])q.pop();
            PLI no=q.top();q.pop();
            int x=no.second;
            if(a[x] > 0 || (L[x] != 0 && R[x] != n + 1))
    		{
                ans -= abs(a[x]);
                a[x] += a[L[x]] + a[R[x]];
                q.push({abs(a[x]), x});
    			remove(L[x]);
    			remove(R[x]);
                cnt--;
            }
        }
        printf("%d", ans);
        return 0;
    }
    
    • 0
      @ 2025-10-8 16:56:02
      #include<bits/stdc++.h>
      using namespace std;
      typedef long long LL;
      typedef pair<LL,int> PLI;
      const int N=1e5+100;
      LL a[N];
      int L[N],R[N],del[N];
      priority_queue<PLI,vector<PLI>,greater<PLI>> q;
      void remove(int x)
      {
          L[R[x]]=L[x];
          R[L[x]]=R[x];
          del[x]=1;
      }
      int main()
      {
          int n,m;scanf("%d%d",&n,&m);
          int k=0;
          for(int i=1;i<=n;i++)
      	{
              LL x;scanf("%lld",&x);
              if(x!=0)
      		{
                  if(!k||a[k]*x<0)a[++k]=x;
                  else a[k]+=x;
              }
          }
          n=k;
          LL ans=0,cnt=0;
          for(int i=1;i<=n;i++)
      	{
              L[i]=i-1,R[i]=i+1;
              q.push({abs(a[i]),i});
              if(a[i]>0)ans+=a[i],cnt++;
          }
          memset(del,0,sizeof(del));
          while(cnt>m)
      	{
              while(del[q.top().second])q.pop();
              PLI no=q.top();q.pop();
              int x=no.second;
              if(a[x]>0||(L[x]!=0&&R[x]!=n+1))
      		{
                  ans-=abs(a[x]);
                  a[x]+=a[L[x]]+a[R[x]];
                  q.push({abs(a[x]),x});
      			remove(L[x]);
      			remove(R[x]);
                  cnt--;
              }
          }
          printf("%d",ans);
          return 0;
      }
      • 1

      *【链表+堆】序列m个连续和最大[CH1812]生日礼物

      信息

      ID
      1304
      时间
      1000ms
      内存
      64MiB
      难度
      6
      标签
      递交数
      99
      已通过
      29
      上传者