【题解】
贪心题。设五元组(mx,pos,l,r1,r2)表示最大值为mx,取得最大值的区间右端点为pos,区间左端点为l,区间右端点的可选区间为[r1,r2]. 每次从堆里拎出最大值,然后把这个区间拆了,因为选了[l,pos]这个区间之后就不能再选它了。我们得往堆了丢俩新的五元组(mx',pos',l,r1,pos-1)以及(mx'',pos'',l,pos+1,r2),至于mx',mx'',pos',pos''的获得,用ST表即可。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<algorithm> 3 #include<cmath> 4 #include<cstring> 5 #define N 1000010 6 #define rg register 7 #define LL long long 8 using namespace std; 9 int n,k,L,R,tot,pos[20][N],f[20][N]; 10 LL ans; 11 struct heap{ 12 LL d,p,l,r1,r2; 13 }h[N]; 14 struct rec{ 15 LL d,p; 16 }; 17 inline int read(){ 18 int k=0,f=1; char c=getchar(); 19 while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar(); 20 while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar(); 21 return k*f; 22 } 23 24 inline void up(int x){ 25 int fa; 26 while((fa=x>>1)&&h[fa].d<h[x].d) swap(h[fa],h[x]),x=fa; 27 } 28 inline void down(int x){ 29 int son; 30 while((son=x<<1)<=tot){ 31 if(son<tot&&h[son+1].d>h[son].d) son++; 32 if(h[son].d>h[x].d) swap(h[son],h[x]),x=son; 33 else return; 34 } 35 } 36 inline rec query(int l,int r){ 37 r=min(r,n); l=max(1,l); 38 int k=log2(r-l+1); rec tmp; tmp.d=tmp.p=0; 39 if(l>r) return tmp; 40 if(f[k][l]>f[k][r-(1<<k)+1]){ 41 tmp.d=f[k][l]; tmp.p=pos[k][l]; 42 return tmp; 43 } 44 else{ 45 tmp.d=f[k][r-(1<<k)+1]; tmp.p=pos[k][r-(1<<k)+1]; 46 return tmp; 47 } 48 } 49 int main(){ 50 memset(f,128,sizeof(f)); f[0][0]=0; 51 n=read(); k=read(); L=read()-1; R=read()-1; 52 for(rg int i=1;i<=n;i++) f[0][i]=read()+f[0][i-1],pos[0][i]=i; 53 for(rg int i=1;i<=log2(n);i++) 54 for(rg int j=1;j<=n-(1<<i)+1;j++) 55 if(f[i-1][j]>f[i-1][j+(1<<(i-1))]) 56 f[i][j]=f[i-1][j],pos[i][j]=pos[i-1][j]; 57 else f[i][j]=f[i-1][j+(1<<(i-1))],pos[i][j]=pos[i-1][j+(1<<(i-1))]; 58 for(rg int i=1;i<=n-L;i++){ 59 rec tmp=query(i+L,i+R); 60 h[++tot]=(heap){tmp.d-f[0][i-1],tmp.p,i,i+L,min(n,i+R)}; 61 up(tot); 62 } 63 for(rg int i=1;i<=k;i++){ 64 heap now=h[1]; 65 ans+=now.d; 66 h[1]=h[tot--]; if(tot) down(1); 67 if(now.p-1>=now.r1){ 68 rec tmp=query(now.r1,now.p-1); 69 h[++tot]=(heap){tmp.d-f[0][now.l-1],tmp.p,now.l,now.r1,now.p-1}; 70 up(tot); 71 } 72 if(now.r2>=now.p+1){ 73 rec tmp=query(now.p+1,now.r2); 74 h[++tot]=(heap){tmp.d-f[0][now.l-1],tmp.p,now.l,now.p+1,now.r2}; 75 up(tot); 76 } 77 } 78 printf("%lld ",ans); 79 return 0; 80 }