bzoj2006

确认右端点后,即可确认左端点的范围,则能确认该右端点能取到的最大值(用前缀和维护sum=s[i]-s[j],s[i]确认,则用ST表维护s[j]最小值)
用优先队列维护,每次取最大值,再将以该右端点为右端点的左右两段区间最值分别放入队列中

#include<cstdio> #include<cctype> #include<queue> #define maxn 500002 #define mp(a,b,c,d) make_pair(make_pair(a,b),make_pair(c,d)) using namespace std; typedef pair<int,int>data; priority_queue<pair<data,data> >q; int n,m,l,r,logg[maxn],f[maxn][20],s[maxn],v[maxn]; long long ans; inline void read(int &x){ char ch=getchar();x=0;int f=1; while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} x*=f; } inline int mn(int a,int b){return s[a]<s[b]?a:b;} inline int query(int a,int b){if(a>b)return -1;int k=logg[b-a+1];return mn(f[a][k],f[b-(1<<k)+1][k]);} int main(){ int c,d,x,a,b,y; read(n);read(m);read(l);read(r); for(int i=2;i<=n;i++)logg[i]=logg[i>>1]+1; for(int i=1;i<=n;i++){f[i][0]=i;read(v[i]);s[i]=v[i]+s[i-1];} for(int j=1;(1<<j)<=n;j++) for(int i=0;i+(1<<j)<=n;i++) f[i][j]=mn(f[i][j-1],f[i+(1<<j-1)][j-1]); for(int i=l;i<=n;i++)q.push(mp(s[i]-s[query(max(i-r,0),i-l)],i,max(i-r,0),i-l)); for(int i=1;i<=m;i++){ data t1=q.top().first,t2=q.top().second; ans+=t1.first;x=t1.second;a=t2.first;b=t2.second;y=query(a,b);q.pop(); c=query(a,y-1);d=query(y+1,b); if(c!=-1)q.push(mp(s[x]-s[c],x,a,y-1)); if(d!=-1)q.push(mp(s[x]-s[d],x,y+1,b)); } printf("%lld",ans); return 0; }

 

 
posted @ 2018-06-09 15:38  lnyzo  阅读(158)  评论(0编辑  收藏  举报