Bzoj2288 birthday gift (line segment tree)

Of course I want to select the largest child segment and the greatest child segment. But if I want to select m, it is not necessarily the best.

Therefore, we provide an option to repent. After selecting a paragraph, I will multiply them by-1 and then select the best one (similar to the idea of network flow)

This can be maintained by using the line segment tree, recording that a range contains the maximum and minimum values of the left and right endpoints (because it must be multiplied by-1) and their endpoint locations.

Then keep searching until the maximum value is <= 0

 1 #include<bits/stdc++.h> 2 #define pa pair<int,int> 3 #define CLR(a,x) memset(a,x,sizeof(a)) 4 #define mp make_pair 5 using namespace std; 6 typedef long long ll; 7 const int maxn=1e5+10,inf=0x3f3f3f3f; 8  9 inline ll rd(){10     ll x=0;char c=getchar();int neg=1;11     while(c<‘0‘||c>‘9‘){if(c==‘-‘) neg=-1;c=getchar();}12     while(c>=‘0‘&&c<=‘9‘) x=x*10+c-‘0‘,c=getchar();13     return x*neg;14 }15 struct Pos{16     int v,l,r;17     Pos (int a=0,int b=0,int c=0){v=a,l=b,r=c;}18 };19 struct Node{20     Pos lmas,rmas,lmis,rmis,ma,mi,sum;21 }tr[maxn<<2];22 23 bool laz[maxn<<2];24 int N,M,v[maxn];25 26 bool operator <(Pos a,Pos b){27     return a.v<b.v;28 }29 Pos operator +(Pos a,Pos b){30     return Pos(a.v+b.v,a.l,b.r);31 }32 Pos operator -(Pos x){return Pos(-x.v,x.l,x.r);}33 34 Node operator +(Node a,Node b){35     Node x;36     x.sum=a.sum+b.sum;37     x.lmas=max(a.lmas,a.sum+b.lmas);38     x.rmas=max(b.rmas,a.rmas+b.sum);39     x.lmis=min(a.lmis,a.sum+b.lmis);40     x.rmis=min(b.rmis,a.rmis+b.sum);41     x.ma=max(a.rmas+b.lmas,max(a.ma,b.ma));42     x.mi=min(a.rmis+b.lmis,min(a.mi,b.mi));43     return x;44 }45 46 inline void build(int p,int l,int r){47     if(l==r){48         tr[p].sum=tr[p].lmas=tr[p].rmas=tr[p].lmis=tr[p].rmis=tr[p].ma=tr[p].mi=Pos(v[l],l,r);49     }else{50         int m=l+r>>1;51         build(p<<1,l,m),build(p<<1|1,m+1,r);52         tr[p]=tr[p<<1]+tr[p<<1|1];53     }54 }55 56 void deal(int p){57     laz[p]^=1;58     Pos t=tr[p].lmas;59     tr[p].lmas=-tr[p].lmis;tr[p].lmis=-t;60     t=tr[p].rmas;61     tr[p].rmas=-tr[p].rmis;tr[p].rmis=-t;62     t=tr[p].ma;63     tr[p].ma=-tr[p].mi;tr[p].mi=-t;64     tr[p].sum=-tr[p].sum;65 }66 67 inline void pushdown(int p){68     if(!laz[p]) return;69     deal(p<<1),deal(p<<1|1);70     laz[p]=0;71 }72 73 inline void rever(int p,int l,int r,int x,int y){74     if(x<=l&&r<=y) deal(p);75     else{76         pushdown(p);77         int m=l+r>>1;78         if(x<=m) rever(p<<1,l,m,x,y);79         if(y>=m+1) rever(p<<1|1,m+1,r,x,y);80         tr[p]=tr[p<<1]+tr[p<<1|1];81     }82 }83 84 int main(){85     //freopen("","r",stdin);86     int i,j,k;87     N=rd(),M=rd();88     for(i=1;i<=N;i++)89         v[i]=rd();90     int ans=0;91     build(1,1,N);92     for(i=1;i<=M;i++){93         Pos x=tr[1].ma;if(x.v<=0) break;94         ans+=x.v;95         rever(1,1,N,x.l,x.r);96     }97     printf("%d\n",ans);98     return 0;99 }

 

Bzoj2288 birthday gift (line segment tree)