題目來源:[NWPU][2014][TRN][13]線段樹第一講 G 題
http://vjudge.net/contest/view.action?cid=50850#problem/G
作者:npufz
題目:
Description
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
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
Sample Input
Sample Output
代碼:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int high1;
typedef struct rootpoint
{
int le,ri;
int mawid;
int mid()
{
return (le+ri)/2;
}
} rootpoi;
rootpoi tree[800050];
int built (int root,int l,int r,int wid )
{
if(l==r)
{
tree[root].le=l;
tree[root].ri=r;
tree[root].mawid=wid;
return 0;
}
if(l!=r)
{
tree[root].le=l;
tree[root].ri=r;
tree[root].mawid=wid;
built(root*2,l,(l+r)/2,wid);
built(root*2+1,(l+r)/2+1,r,wid);
}
return 0;
}
int zhantie(int root,int wid1)
{
if(tree[root].mawid<wid1) return 0;
if(tree[root].le==tree[root].ri&&tree[root].mawid>=wid1)
{
high1=tree[root].le;
tree[root].mawid=tree[root].mawid-wid1;
return 0;
}
else if(tree[root*2].mawid>=wid1)
{
zhantie(root*2,wid1);
}
else
{
zhantie(root*2+1,wid1);
}
tree[root].mawid=max(tree[root*2].mawid,tree[root*2+1].mawid);
return 0;
}
int main()
{
int n,hi,wid,i,j,k,realhi;
while(~scanf("%d%d%d",&hi,&wid,&n))
{
realhi=min(hi,n);
built (1,1,realhi,wid);
for(i=0;i<n;i++)
{ int wid1;
scanf("%d",&wid1);
high1=-1;
zhantie(1,wid1);
printf("%d\n",high1);
}
}
return 0;
}
反思:
線段樹的構造主要是維護好子節點的變化對根節點的影響,看準輸入輸出很重要