This question of the Internet seems to be little problem solution? Write a bar
I traced back to the last semester just learning segment tree that day with this question of origins. . .
At that time the tree line in front of the thematic question are some of the board after a short while on the water on, and then you see the last question of it: Shan Hai Jing
That morning, I did my best, but there is no harvest.
Later in the afternoon I continued anus or anal does not move, but mikufun Great God, at the stage of just learning segment tree, and now I have to write this tract problems 30min to the A!
He was the NB Code:
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 struct tree{ 5 int l,r,da,dam,dal,dar; 6 int lme,lmr,le,lr; 7 }t[1000000]; 8 void build(int l1,int r1,int k){ 9 t[k].l=l1,t[k].r=r1; 10 if(l1==r1){ 11 scanf("%d",&t[k].da); 12 t[k].dal=t[k].dam=t[k].dar=t[k].da; 13 t[k].le=t[k].lme=t[k].lmr=t[k].lr=l1; 14 return; 15 } 16 int mid=(l1+r1)/2; 17 build(l1,mid,k*2); 18 build(mid+1,r1,k*2+1); 19 t[k].da=t[k*2].da+t[k*2+1].da; 20 if(t[k*2].dal>=t[k*2].da+t[k*2+1].dal){ 21 t[k].dal=t[k*2].dal; 22 t[k].le=t[k*2].le; 23 } 24 else{ 25 t[k].dal=t[k*2].da+t[k*2+1].dal; 26 t[k].le=t[k*2+1].le; 27 } 28 if(t[k*2+1].dar>t[k*2+1].da+t[k*2].dar){ 29 t[k].dar=t[k*2+1].dar; 30 t[k].lr=t[k*2+1].lr; 31 } 32 else{ 33 t[k].dar=t[k*2+1].da+t[k*2].dar; 34 t[k].lr=t[k*2].lr; 35 } 36 if(max(t[k*2].dam,max(t[k*2+1].dam,t[k*2].dar+t[k*2+1].dal))==t[k*2].dam){ 37 t[k].dam=t[k*2].dam; 38 t[k].lme=t[k*2].lme;t[k].lmr=t[k*2].lmr; 39 } 40 else if(max(t[k*2].dam,max(t[k*2+1].dam,t[k*2].dar+t[k*2+1].dal))==t[k*2+1].dam){ 41 t[k].dam=t[k*2+1].dam; 42 t[k].lme=t[k*2+1].lme;t[k].lmr=t[k*2+1].lmr; 43 } 44 else if(max(t[k*2].dam,max(t[k*2+1].dam,t[k*2].dar+t[k*2+1].dal))==t[k*2].dar+t[k*2+1].dal){ 45 t[k].dam=t[k*2].dar+t[k*2+1].dal; 46 t[k].lme=t[k*2].lr;t[k].lmr=t[k*2+1].le; 47 } 48 } 49 inline int sea_qu(int l1,int r1,int k,int x,int &ll,int &rr){ 50 if(l1<=t[k].l&&t[k].r<=r1){ 51 if(!x){ 52 ll=t[k].lme;rr=t[k].lmr; 53 return t[k].dam; 54 } 55 if(x==1){ 56 ll=t[k].lr;rr=t[k].r; 57 return t[k].dar; 58 } 59 if(x==2){ 60 ll=t[k].l;rr=t[k].le; 61 return t[k].dal; 62 } 63 } 64 int mid=(t[k].l+t[k].r)/2,ans,a1,a2,a3,a4,a5,a6; 65 int a,b,c,d,e,g,h,z; 66 if(x==1){ 67 a1=sea_qu(l1,r1,k*2+1,1,a,b); 68 ll=a;rr=t[k].r; 69 if(l1<=mid){ 70 a5=sea_qu(l1,r1,k*2,1,b,c)+t[k*2+1].da; 71 if(a5>a1){ 72 ll=b,rr=t[k].r; 73 return a5; 74 } 75 } 76 return a1; 77 } 78 if(x==2){ 79 a2=sea_qu(l1,r1,k*2,2,z,c); 80 rr=c;ll=t[k].l; 81 if(r1>mid){ 82 a6=sea_qu(l1,r1,k*2+1,2,b,z)+t[k*2].da; 83 if(a6>a2){ 84 ll=t[k].l,rr=z; 85 return a6; 86 } 87 } 88 return a2; 89 } 90 if(!x){ 91 if(r1<=mid){ 92 ans=sea_qu(l1,r1,k*2,0,ll,rr); 93 return ans; 94 } 95 if(l1>mid){ 96 ans=sea_qu(l1,r1,k*2+1,0,ll,rr); 97 return ans; 98 } 99 a1=sea_qu(l1,r1,k*2,1,a,b); 100 a2=sea_qu(l1,r1,k*2+1,2,b,c); 101 a3=sea_qu(l1,r1,k*2,0,h,d); 102 a4=sea_qu(l1,r1,k*2+1,0,e,g); 103 if(max(a1+a2,max(a3,a4))==a3){ 104 ll=h,rr=d; 105 return a3; 106 } 107 if(max(a1+a2,max(a3,a4))==a1+a2){ 108 ll=a; 109 rr=c; 110 return a1+a2; 111 } 112 if(max(a1+a2,max(a3,a4))==a4){ 113 ll=e,rr=g; 114 return a4; 115 } 116 } 117 } 118 int main(){ 119 int n,m,x,y,ll,rr; 120 scanf("%d%d",&n,&m); 121 build(1,n,1); 122 for(int i=1;i<=m;i++){ 123 scanf("%d%d",&x,&y); 124 int ans=sea_qu(x,y,1,0,ll,rr); 125 cout<<ll<<" "<<rr<<" "<<ans<<endl; 126 } 127 return 0; 128 }
This is at the time, which is no problem when play 50 lines + ah.
I did see someone up A shocked.
I had to ask mikufun, then he said:. "Yourself, think again."
A later I always wanted out of this question, you can always have some reason to stop the pace I stumble forward.
But now I finally A, and easily relieved.
I remember watching Laolv only mikufunA This question encourages us:. "Nothing, as long as the knowledge is not lost, these problems are very simple to see you later."
But I was like: "so hard question, I'm afraid I'll never do wrong." I did not expect ah, only a few months, the change has been so big.
OI change you and me.
Solution: maximum segment tree maintenance starting left end point, maximum, maximum interval start right end point, as well as their endpoints.
In sub-tree to merge his father, his father left the maximum starting point can be a maximum starting from the left point he left his son, his son left + right maximum value of all the starting left end point update.
Similarly to the maximum point on the right end of the father, the maximum range. But the maximum range can also be updated by the maximum value of the right + left endpoint son left his son the right point.
Overall the details with a lot of good attention to it lexicographical problem. In fact, the order of a few adjustments Bale.
When merging merge my subsequent query for convenience, be combined with the selected structure, because since when find find [a, b] is divided into sections log intervals, therefore equivalent to the father incorporated subtree .
1 #include<bits/stdc++.h> 2 #define N 100005 3 #define Inf 0x7f7f7f7f 4 #define lch k<<1 5 #define rch k<<1|1 6 inline int read(){ 7 int x=0,f=1;char ch=getchar(); 8 while(!isdigit(ch))f=(ch=='-'?-1:1),ch=getchar(); 9 while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); 10 return x*f; 11 } 12 struct node{int l,r,mx,lmx,lp,rmx,rp,smx,slp,srp;}tr[N<<2]; 13 int n,m; 14 node merge(const node &t1,const node &t2){ 15 node ret;ret.lmx=ret.rmx=ret.smx=-Inf; 16 ret.l=t1.l,ret.r=t2.r;ret.mx=t1.mx+t2.mx; 17 ret.lmx=t1.lmx,ret.lp=t1.lp; 18 ret.rmx=t2.rmx,ret.rp=t2.rp; 19 if(t1.smx> = t2.smx) ret.smx = t1.smx, ret.slp = t1.slp, ret.srp = t1.srp; 20 IF (t1.rmx + t2.lmx> ret.smx) ret.smx = t1.rmx + t2.lmx, ret.slp = t1.rp, ret.srp = t2.lp; 21 IF (t1.rmx + t2.lmx == ret.smx && ((t1.rp <ret.slp) || (t1.rp == ret.slp && t2.lp <ret.srp))) ret.smx = + t1.rmx t2.lmx, ret.slp = t1.rp, ret.srp = t2.lp; 22 IF (t2.smx> ret.smx) ret.smx = t2.smx, ret.slp = t2.slp, ret.srp = t2.srp; 23 IF (t1.mx + t2.lmx> ret.lmx) ret.lmx = t1.mx + t2.lmx, ret.lp = t2.lp; 24 IF (t2.mx + t1.rmx> ret.rmx) ret.rmx = t2.mx + t1.rmx, ret.rp = t1.rp; 25 IF (ret.smx <= ret.lmx) ret.slp = ret.l, ret.srp =ret.lp; 26 if(ret.smx<ret.rmx) ret.slp=ret.rp,ret.srp=ret.r; 27 return ret; 28 } 29 void build(int k,int l,int r){ 30 tr[k].l=l,tr[k].r=r; 31 if(l==r){tr[k].mx=tr[k].lmx=tr[k].rmx=tr[k].smx=read();tr[k].lp=tr[k].rp=tr[k].slp=tr[k].srp=l;return;} 32 const int mid=(l+r)>>1; 33 build(lch,l,mid),build(rch,mid+1,r); 34 tr[k]=merge(tr[lch],tr[rch]); 35 } 36 node find(int k,int l,int r){ 37 if(tr[k].l>=l&&tr[k].r<=r) return tr[k]; 38 node ret; 39 if(l<=tr[lch].r&&r>=tr[rch].l) ret=merge(find(lch,l,r),find(rch,l,r)); 40 else if(r<=tr[lch].r) ret=find(lch,l,r); 41 else if(l>=tr[rch].l) ret=find(rch,l,r); 42 return ret; 43 } 44 int main(){ 45 n=read(),m=read(); 46 build(1,1,n); 47 int a,b; 48 for(int i=1;i<=m;++i){ 49 a=read(),b=read(); 50 node ret=find(1,a,b); 51 printf("%d %d %d\n",ret.slp,ret.srp,ret.smx); 52 } 53 }