Billboard
Time Limit: 20000/8000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12249Accepted Submission(s): 5380
Problem Description
At the entrance to the university,there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions,changes in the dining room menu,and other important information.
On September 1,the billboard was empty. One by one,the announcements started being put on the billboard.
Each announcement is a stripe of paper of unit height. More specifically,the i-th announcement is a rectangle of size 1 * wi.
When someone puts a new announcement on the billboard,she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.
If there is no valid location for a new announcement,it is not put on the billboard (that's why some programming contests have no participants from this university).
Given the sizes of the billboard and the announcements,your task is to find the numbers of rows in which the announcements are placed.
On September 1,the billboard was empty. One by one,the announcements started being put on the billboard.
Each announcement is a stripe of paper of unit height. More specifically,the i-th announcement is a rectangle of size 1 * wi.
When someone puts a new announcement on the billboard,she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.
If there is no valid location for a new announcement,it is not put on the billboard (that's why some programming contests have no participants from this university).
Given the sizes of the billboard and the announcements,your task is to find the numbers of rows in which the announcements are placed.
Input
There are multiple cases (no more than 40 cases).
The first line of the input file contains three integer numbers,h,w,and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.
Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.
The first line of the input file contains three integer numbers,h,w,and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.
Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.
Output
For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h,starting with the top row. If an announcement can't be put on the billboard,output "-1" for this announcement.
Sample Input
3 5 5 2 4 3 3 3
Sample Output
1 2 1 3 -1
思路:
想象成一个有h层的架子,每层的高度都是1,长度都是w,现在有n个木条,高度都是1,宽度是输入信息,要放入架子,尽可能地往上放,如果能放入,输出层数,放不下,就输出-1
就是说我要随时知道每层的剩余长度,拿到一根木条,找到能放入的最上层,放入后,更新该层的剩余宽度...所以想到了线段树...
构建空树,叶子节点都是w,表示每层的初始宽度都是w,非叶子结点存储的是子节点的最大值,那么根节点就是当前所有层数剩余宽度的最大值
每次拿到一个木条,如果宽度小于等于根节点,说明能放入这个架子,否则直接输出-1
能放入的前提下,每次跟所到节点的左子节点比较,如果小于等于,就往左,反之往右,直到叶子节点,更新剩余宽度,输出层数...
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <iostream> #include <algorithm> using namespace std; const int MAXN=200100; int num[MAXN<<2]; void PushUp(int ret) { num[ret]=max(num[ret<<1],num[ret<<1|1]); } void BuildTree(int w,int left,int right,int ret) { num[ret]=w; if(left==right) return; int mid=(left+right)>>1; BuildTree(w,left,mid,ret<<1); BuildTree(w,mid+1,right,ret<<1|1); } int Query(int len,int ret) { if(num[ret]<len) return -1; if(left==right) { num[ret]-=len; return left; } int mid=(left+right)>>1; int ans; if(num[ret<<1]>=len) ans=Query(len,ret<<1); else ans=Query(len,ret<<1|1); PushUp(ret); return ans; } int main() { freopen("2795.in","r",stdin); // freopen("2795.out","w",stdout); int h,n; while(scanf("%d%d%d",&h,&w,&n)!=EOF) { if(h>n) h=n; BuildTree(w,1,1); int len; while(n--) { scanf("%d",&len); printf("%d\n",Query(len,1)); } } return 0; } /* #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <iostream> #include <algorithm> using namespace std; const int MAXN=200100; int num[MAXN<<2]; void PushUp(int ret) { num[ret]=max(num[ret<<1],ret<<1|1); } void Update(int target,int reduce,int ret) { if(left==right) { num[ret]-=reduce; return ; } int mid=(left+right)>>1; if(target<=mid) Update(target,reduce,ret<<1); if(mid<target) Update(target,ret<<1|1); PushUp(ret); } int Query(int len,int ret) { if(num[ret]<len) return -1; if(left==right) return left; int mid=(left+right)>>1; if(num[ret<<1]>=len) return Query(len,ret<<1); else return Query(len,ret<<1|1); } int main() { freopen("2795.in",1); int len,ans; while(n--) { scanf("%d",&len); ans=Query(len,1); if(-1 != ans) Update(ans,len,1); printf("%d\n",ans); } } return 0; } */
