The machine AC submitted RE ......
Portal: QAQQAQ
Meaning of the questions: they look
Ideas: This question is said to do quickly with RMQ, but this question is a segment tree
Segment tree maintenance period interval in the left-most, right-most years, whether there is a gap and a maximum range
This question is a major difficulty discussing classification, divided into the following situations:
1.AB not determined, the output maybe
2.A determine B uncertainty:
If A is the A to B, the maximum output maybe
Otherwise false output
3.A uncertain B determines:
If B is lower_bound (A) B to the maximum, the output maybe
Otherwise false output
4.A OK B OK
If B is A + 1 B to the maximum, the maximum of A to A is B, with no gaps between the true output
If the above conditions are satisfied, but there is a gap, the output maybe
Otherwise false output
In the code implementation, I did not at first consider whether equal before the "strictly less than" the condition that the maximum value is equal to B on the line, but there is no judgment
Then added some special judge to prevent the cross-border segment tree own metaphysics not stop. . . (Recently code implementation capacity should be improved ah ...)
Code:
#include<bits/stdc++.h> using namespace std; const int N=100005; int n; struct node { int ly,ry,max_r,Gap; }tree[N*4]; int rain[N],year[N]; void push_up(node& fa,node ls,node rs) { fa.ly=ls.ly; fa.ry=rs.ry; fa.max_r=max(ls.max_r,rs.max_r); fa.Gap=0; if(ls.Gap||rs.Gap) fa.Gap=1; if(ls.ry+1<rs.ly) fa.Gap=1; } void build(int x,int l,int r) { if(l==r) { tree[x].ly=year[l]; tree[x].ry=year[l]; tree[x].max_r=rain[l]; tree[x].Gap=0; return; } int mid=(l+r)>>1; build(x+x,l,mid); build(x+x+1,mid+1,r); push_up(tree[x],tree[x+x],tree[x+x+1]); } node query(int x,int l,int r,int L,int R) { node ret; if(L<=l&&R>=r) return tree[x]; int mid=(l+r)>>1; if(mid<L) return query(x+x+1,mid+1,r,L,R); if(mid>=R) return query(x+x,l,mid,L,R); node ls=query(x+x,l,mid,L,R); node rs=query(x+x+1,mid+1,r,L,R); push_up(ret,ls,rs); return ret; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d%d",&year[i],&rain[i]); build(1,1,n); int q; scanf("%d",&q); while(q--) { int x,y,xx,yy,bl=0; scanf("%d%d",&xx,&yy); if(yy<xx) { puts("false"); continue; } if(xx==yy) { puts("maybe"); continue; } x=lower_bound(year+1,year+n+1,xx)-year; y=lower_bound(year+1,year+n+1,yy)-year; if(xx!=year[x]&&yy!=year[y]) puts("maybe"); else if(xx!=year[x]) { if(y==1||x==y) puts("maybe"); else { node n1=query(1,1,n,x,y); node n2=query(1,1,n,x,y-1); //中间降雨量必须严格小于！ if(n1.max_r==rain[y]&&n2.max_r<rain[y]) puts("maybe"); else puts("false"); } } else if(yy!=year[y]) { if(x==n||x==y-1) puts("maybe"); else { node n1=query(1,1,n,x,y-1); node n2=query(1,1,n,x+1,y-1); if(n1.max_r==rain[x]&&n2.max_r<rain[x]) puts("maybe"); else puts("false"); } } else { if(x+1==y) { if(rain[x]<rain[y]) puts("false"); else if(year[x]+1==year[y]) puts("true"); else puts("Maybe " ); continue ; // not written before continue ah ah ... } // Laid determination prevented segment tree bounds Node N1 = Query ( . 1 , . 1 , n-, X, Y); Node N2 = Query ( . 1 , . 1 , n-, X + . 1 , Y); Node N3 = Query ( . 1 , . 1 , n-, X + . 1 , Y- . 1 ); IF (n2.max_r == Rain [Y] && Rain [Y] <= Rain [X] n3.max_r && < Rain [Y]) { IF (n1.Gap) the puts ( " Maybe "); else puts("true"); } else puts("false"); } } return 0; }