Test instructions: N squares, side length is s[i], oblique 45 degrees in order to lay on the axis, as far as possible to the left, but not with any previous intersection, ask from the top down, which squares are visible.
Solution: We first expand the edge length to sqrt (2) times the edge length, so that you can directly perform integer operations.
And then in two steps:
1. Find out all the b[i]
2. Then the interval coverage decision
The first step: when we know all the previous b[j], B[i] is one of the leftmost positions that does not intersect with any of the preceding. To be satisfied with the front is not intersect and the most left, we have: b[i] = max (B[i],b[j]+2*min (S[I],S[J)), according to, push can be drawn.
The second step: the l,r of all squares have been calculated, and then n^2 to update the l,r of each square as visible l,r:
1. If s[i] > S[j] (j=1~i-1), and Seg[i]. L < Seg[j]. R, then seg[j]. R = Seg[i]. L
2. If s[i] < s[j]. , and Seg[i]. L < Seg[j]. R, then seg[i]. L = Seg[j]. R
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Code:
#include <iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespacestd;//Data Segmentints[ -],b[ -];structseg{intL,r,len;} seg[ -];//Data EndsintMain () {intn,i,j; DoubleGen2 = sqrt (2.0); while(SCANF ("%d", &n)!=eof &&N) { for(i=1; i<=n;i++) scanf ("%d",&R[i]); b[1] = s[1]; for(i=2; i<=n;i++) {B[i]= b[i-1]; for(j=i-1; j>=1; j--) B[i]= Max (B[i],b[j] +2*min (s[i],s[j])); } for(i=1; i<=n;i++) {Seg[i]. L= B[i]-S[i]; Seg[i]. R= B[i] +S[i]; Seg[i]. LEN=S[i]; } for(i=1; i<=n;i++) { for(j=1; j<i;j++) { if(Seg[i]. LEN > Seg[j]. LEN && Seg[i]. L <Seg[j]. R) Seg[j]. R=Seg[i]. L if(Seg[i]. LEN < Seg[j]. LEN && Seg[i]. L <Seg[j]. R) Seg[i]. L=Seg[j]. R } } intFIR =1; for(i=1; i<=n;i++) { if(Seg[i]. L <Seg[i]. R) {if(FIR) cout<<i, fir =0; Elsecout<<" "<<i; }} puts (""); } return 0;}
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POJ 3347 Kadj Squares--extended length, geometry, segment coverage