51nod1053 最大M子段和 V2

$n \leq 50000$的序列，问选不超过$m \leq 50000$个区间使得和最大。

如果正数区间总数比$m$小那肯定全选。否则有两种方式减少区间数量：丢掉一个正区间；补一个负区间连接两个正区间。贪心即可。

先把左右端的负数去掉，然后把正区间和负区间处理出来。优先队列维护区间值，然后开个链表模拟合并（删左右，改自己）。注意删右边时调整右端点。

 //#include<iostream>
 #include<cstring>
 #include<cstdio>
 //#include<time.h>
 //#include<complex>
 //#include<set>
 #include<queue>
 //#include<vector>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;
 
 #define LL long long
 int qread()
 {
     char c; int s=,f=; while ((c=getchar())<'' || c>'') (c=='-') && (f=-);
     do s=s*+c-''; while ((c=getchar())>='' && c<=''); return s*f;
 }
 
 //Pay attention to '-' , LL and double of qread!!!!
 
 int n,m;
 #define maxn 50011
 #define LL long long
 
 LL Abs(LL x) {return x>?x:-x;}
 
 int a[maxn]; LL sum[maxn];
 int b[maxn],lb,ll[maxn],rr[maxn]; bool vis[maxn];
 struct qnode
 {
     LL v; int id;
     bool operator > (const qnode &b) const {return v>b.v;}
 };
 priority_queue<qnode,vector<qnode>,greater<qnode> > q;
 int main()
 {
     n=qread(); m=qread();
     for (int i=;i<=n;i++) a[i]=qread();
     {
         int L=,R=n; while (a[L]<=) L++; while (a[R]<=) R--;
         for (int i=L,j=;i<=R;i++,j++) a[j]=a[i];
         n=R-L+; a[n+]=;
     }
     for (int i=;i<=n;i++) sum[i]=sum[i-]+a[i];
 
     LL ans=;
     int cnt=;
     {
         LL tmp=a[],last=;
         for (int i=;i<=n+;i++)
         {
             if (tmp> && a[i]<=)
             {
                 b[++lb]=last;
                 q.push((qnode){tmp,lb});
                 ans+=tmp; cnt++; tmp=a[i]; last=i;
             }
             else if (tmp> && a[i]>) tmp+=a[i];
             else if (tmp<= && a[i]>)
             {
                 b[++lb]=last;
                 q.push((qnode){-tmp,lb});
                 tmp=a[i]; last=i;
             }
             else tmp+=a[i];
         }
     }
 
     for (int i=;i<=lb+;i++) ll[i]=i-,rr[i]=i+; b[lb+]=n+;
     if (cnt<=m) {printf("%lld\n",ans); return ;}
     while (m<cnt--)
     {
         while (vis[q.top().id]) q.pop();
         ans-=q.top().v; int now=q.top().id; q.pop();
 
         if (ll[now]==)
         {
             vis[rr[now]]=;
             vis[now]=;
             int u=rr[now]; ll[rr[now]]=ll[now]; rr[ll[now]]=rr[now];
             ll[rr[u]]=ll[u]; rr[ll[u]]=rr[u];
         }
         else if (rr[now]==lb+)
         {
             vis[ll[now]]=;
             vis[now]=;
             int u=ll[now]; ll[rr[now]]=ll[now]; rr[ll[now]]=rr[now];
             ll[rr[u]]=ll[u]; rr[ll[u]]=rr[u]; b[lb+]=b[u];
         }
         else
         {
             int L=b[ll[now]],R=b[rr[rr[now]]]-;
             vis[ll[now]]=; vis[rr[now]]=; int u=ll[now],v=rr[now];
             b[now]=L; q.push((qnode){Abs(sum[R]-sum[L-]),now});
             ll[rr[u]]=ll[u]; rr[ll[u]]=rr[u];
             ll[rr[v]]=ll[v]; rr[ll[v]]=rr[v];
         }
     }
 
     printf("%lld\n",ans);
     return ;
 }