http://poj.org/problem?id=3667
The hotel has n room number 1 to n are vacant, then M asked, when the input of the first 1, the representative to live in x consecutive rooms, enter the minimum number of room number, if not
Output 0. When the first number is 2, the room from the X number to the Y number is empty, no output
Related to the interval to think of a line tree can be solved, just like the color of the room to live with the number of non-living mark on the line, the problem is that the length of the interval is not
The specific interval, the processing is more troublesome to find the appropriate interval, so the introduction of LEN1 to indicate the maximum length available for a range, Len2 represents the maximum length available from the left end of the interval,
The LEN3 represents the maximum length available from the right end of the interval, and the last update maintains
Code
1#include <iostream>2#include <cstdio>3 using namespacestd;4 structPoint {5 intL,r;6 intLen1,len2,len3;7 intMark;//lazy Judge8 };9Point tree[50001*4];Ten voidBuildintIintLeftintRight ) One { ATree[i].l=left,tree[i].r=Right ; -tree[i].len1=tree[i].len2=tree[i].len3=tree[i].r-tree[i].l+1;//The length starts with the interval length -tree[i].mark=0; the if(Left==right)return; - intMid= (left+right)/2; -Build (i*2, left,mid); -Build (i*2+1, mid+1, right); + } - voidUpdateintIintLeftintRightintval) + { A if(tree[i].l==left&&tree[i].r==Right ) at { - inte; -tree[i].mark=Val; - if(Val) e=0and//if occupied, the interval length is 0. - Elsee=tree[i].r-tree[i].l+1; -tree[i].len1=tree[i].len2=tree[i].len3=e; in return; - } to if(tree[i].mark!=-1) + { - intq=Tree[i].mark,e,r; thetree[i*2].mark=tree[i*2+1].mark=Tree[i].mark; *tree[i].mark=-1; $ if(q) e=0, r=0;Panax Notoginseng Else - { thee=tree[i*2].r-tree[i*2].l+1; +r=tree[i*2+1].r-tree[i*2+1].l+1; A } thetree[i*2].len1=tree[i*2].len2=tree[i*2].len3=e; +tree[i*2+1].len1=tree[i*2+1].len2=tree[i*2+1].len3=R; - } $ intMid= (TREE[I].R+TREE[I].L)/2; $ if(left>mid) Update (i*2+1, left,right,val); - Else if(right<=mid) Update (i*2, left,right,val); - Else the { -Update (i*2, left,mid,val);WuyiUpdate (i*2+1, mid+1, right,val); the } - intTMP = MAX (tree[i*2].len1,tree[i*2+1].len1); WuTree[i].len1=max (tmp,tree[i*2].len3 + tree[i*2+1].len2); -tree[i].len2=tree[i*2].len2; Abouttree[i].len3=tree[i*2+1].len3; $ if(tree[i*2].len1 = = tree[i*2].r-tree[i*2].l+1)//To maintain the len2 and Len3 of the sub-range -tree[i].len2+=tree[i*2+1].len2; - if(tree[i*2+1].len1 = = tree[i*2+1].r-tree[i*2+1].l+1) -tree[i].len3+=tree[i*2].len3; A return ; + } the intFindintIintlen) - { $ if(tree[i].l==tree[i].r&&len==1)returnTREE[I].L; the if(tree[i].mark!=-1) the { the intq=Tree[i].mark,e,r; thetree[i*2].mark=tree[i*2+1].mark=Tree[i].mark; -tree[i].mark=-1; in if(q) e=0, r=0; the Else the { Aboute=tree[i*2].r-tree[i*2].l+1; ther=tree[i*2+1].r-tree[i*2+1].l+1; the } thetree[i*2].len1=tree[i*2].len2=tree[i*2].len3=e; +tree[i*2+1].len1=tree[i*2+1].len2=tree[i*2+1].len3=R; - } the if(tree[i*2].len1>=len)returnFind (i*2, Len);Bayi Else if(tree[i*2].len3+tree[i*2+1].len2>=len)return(tree[i*2].r-tree[i*2].len3+1); the Else if(tree[i*2+1].len1>=len)returnFind (i*2+1, Len); the Else return 0; - } - intMain () the { the intn,m,x,y,z,w; the while(~SCANF ("%d%d",&n,&m)) the { -Build1,1, n); the while(m--) the { thescanf"%d",&x);94 if(x==1) the { thescanf"%d",&y); theW=find (1, y);98printf"%d\n", W); About if(w) Update (1, w,w+y-1,1); - }101 if(x==2)102 {103scanf"%d%d",&y,&z);104Update1, y,y+z-1,0); the }106 }107 }108 return 0;109}
POJ 3667 (segment tree interval merger) open the room.