Billboard
Time Limit:8000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 2795
Appoint description:
System Crawler (2015-04-10)
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 5
2
4
3
3
3
Sample Output
1
2
1
3
-1
解题思路：给出的h和w都是10^9。开不了这么大数数组。但是给的需要张贴的通告n却只有10^5，也就可以开这么大的数组了。各个叶子结点存的值为通告栏该行的最大宽度，而分支结点存的是子结点中的最大值。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=222222; //最多贴n个通告，所以就算h很大，只要满足n个通告就行了
int maxv[maxn*4];
void PushUP(int rt){
maxv[rt]=max(maxv[rt*2],maxv[rt*2+1]);
}
void build(int rt,int L,int R,int wid){
maxv[rt]=wid;
if(L==R){
return ;
}
build(lson,wid);
build(rson,wid);
}
int query(int rt,int L,int R,int len){
if(L==R){
maxv[rt]-=len;
return L;
}
int ret=0;
if(maxv[rt*2]>=len){
ret= query(lson,len);
}else {ret= query(rson,len);}
PushUP(rt);
return ret;
}
int main(){
int n,h,w;
while(scanf("%d%d%d",&h,&w,&n)!=EOF){
if(n<h){
h=n;
}
build(1,1,h,w);
for(int i=0;i<n;i++){
int len;
scanf("%d",&len);
if(maxv[1]<len)
printf("-1\n");
else
printf("%d\n",query(1,1,h,len)) ;
}
}
return 0;
}