HDU2795--Billboard

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.

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.

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 524333

Sample Output

1213-1

 

/*顶多有20万张海报也就是最多也就是需要用到20W行key表示这区间的行中最大剩余长度。*/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <set>#include <map>using namespace std;#define maxn 200008//高度和张数小的那个做N就行了int n,w,h,ans;inline int max(int a,int b){return a>b?a:b;}inline int min(int a,int b){return a>b?b:a;}struct ST{int l,r,key;//表示这一个区间最大的数}st[4*maxn];void buildtree(int id,int l,int r){st[id].l=l;st[id].r=r;if(l==r){st[id].key=w;return;}int mid=(l+r)>>1;buildtree(2*id,l,mid);buildtree(2*id+1,mid+1,r);st[id].key=w;}void update(int id,int ckey){if(st[id].l==st[id].r){if(st[id].key>=ckey){st[id].key-=ckey;ans=st[id].l;}else ans=-1;return;}if(st[2*id].key>=ckey){update(2*id,ckey);st[id].key=max(st[2*id].key,st[2*id+1].key);return;}if(st[2*id+1].key>=ckey){update(2*id+1,ckey);st[id].key=max(st[2*id].key,st[2*id+1].key);return ;}ans=-1;return;}int main(){while(scanf("%d%d%d",&h,&w,&n)==3){ans=-1;if(h>n)buildtree(1,1,n);else buildtree(1,1,h);int len;for(int i=1;i<=n;i++){scanf("%d",&len);update(1,len);printf("%d\n",ans);}}return 0;}

 代码风格更新后：

#include <iostream>#include <cstring>#include <cstdio>using namespace std;#define lson 2*id,l,mid#define rson 2*id+1,mid+1,r#define maxn 200008int h,w,n,pos;//初始的宽度struct ST{int l,r,key;//key用来存该区间的行中最长剩余长度}st[4*maxn];inline int max(int a,int b){return a>b?a:b;}void PushUp(int id){st[id].key=max(st[2*id].key,st[2*id+1].key);}void buildtree(int id,int l,int r){st[id].l=l;st[id].r=r;if(l==r){st[id].key=w;return;}int mid=(l+r)>>1;buildtree(lson);buildtree(rson);st[id].key=w;}void update(int id,int cost){if(st[id].l==st[id].r){if(st[id].key>=cost){pos=st[id].l;st[id].key-=cost;}else pos=-1;return;}if(st[2*id].key>=cost){update(2*id,cost);PushUp(id);return;}if(st[2*id+1].key>=cost){update(2*id+1,cost);PushUp(id);return;}pos=-1;}int main(){while(scanf("%d%d%d",&h,&w,&n)==3){buildtree(1,1,min(h,n));int key;for(int i=1;i<=n;i++){scanf("%d",&key);update(1,key);printf("%d\n",pos);}}return 0;}