【POJ Challenge】生日礼物 加强m子段和

还是链表跟二叉堆的双映射

//#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
using namespace std;         //

#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
	ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{ll x;ll y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
const int dx[4]={0,0,-1,1};
const int dy[4]={1,-1,0,0};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3f;

const ll mod=1e9+7;;
const int maxn=2e5+3;

int ans;

struct heap{
	int v,x,av;
}h[maxn];int sz;
int n,m,a[maxn];//read
int cnt,v[maxn];//正块个数
int tot;		//选择数

int pre[maxn],nxt[maxn],pos[maxn];

void pushup(int x){
	while(h[x].av<h[x>>1].av){
		pos[h[x>>1].x]=x;
		swap(h[x],h[x>>1]);
		x=x>>1;
	}
	pos[h[x].x]=x;
}
void push(int v,int x){
	sz++;h[sz].v=v;h[sz].x=x;h[sz].av=abs(v);
	pos[x]=sz;
	pushup(sz);
}
void pushdown(int x){
	int to;
	while((x<<1)<=sz){
		to=(x<<1);
		if(to<sz && h[to].av>h[to+1].av)to++;
		if(h[x].av>h[to].av){
			pos[h[to].x]=x;
			swap(h[to],h[x]);
			x=to;
		}else break;
	}
	pos[h[x].x]=x;
}
void del(int x){
	h[x].av=inf;
	pushdown(x);
}

void solve(){
	int a,b;heap k;
	for(int i=1;i<=cnt;i++)push(v[i],i);
	while(tot>m){
		tot--;
		k=h[1];
		if(pre[k.x]==-1){
			if(k.v>0){ans-=k.v;del(1);}
			else{del(1);tot++;}
			pre[nxt[k.x]]=-1;
		}else if(nxt[k.x]==-1){
			if(k.v>0){ans-=k.v;del(1);}
			else{del(1);tot++;}
			nxt[pre[k.x]]=-1;
		}else{
			ans-=k.av;
			a=nxt[k.x];b=pre[k.x];
			pre[k.x]=pre[b];nxt[pre[b]]=k.x;
			nxt[k.x]=nxt[a];pre[nxt[a]]=k.x;
			h[1].av=h[pos[a]].av+h[pos[b]].av-h[1].av;
			h[1].v+=h[pos[a]].v+h[pos[b]].v;
			pushdown(1);
			del(pos[a]);del(pos[b]);
		}
	}
	printf("%d\n",ans);
}

int main(){
	scanf("%d%d",&n,&m);
	int tmp=0;
	for(int i=1;i<=n;i++){
		int x;scanf("%d",&x);
		a[++tmp]=x;
		if(a[tmp]==0)tmp--;
	}
	v[++cnt]=a[1];
	for(int i=2;i<=tmp;i++){
		if(a[i]*a[i-1]>0)v[cnt]+=a[i];
		else v[++cnt]=a[i];
	}
	for(int i=1;i<=cnt;i++)if(v[i]>0){ans+=v[i];tot++;}
	for(int i=1;i<=cnt;i++){nxt[i]=i+1;pre[i]=i-1;}
	nxt[cnt]=-1;pre[1]=-1;
	solve();
}