Billboard
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=2795Description
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
2
1
3
-1
HINT
题意
有一个w*h的版，然后让你贴高度为1长度为x的海报，海报贴的方法是，优先贴最高的，然后再贴最左边的
题解：
线段树单点更新
主要问题就是建树的时候，需要稍微注意一下，当h>n时，另h=n就好了。
Debug:
hdu上，ONLINE_JUDGE打错了为什是WA，以前都是RE。还是so vegetable。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 500050 4 int n,w,h; 5 struct Tree{int l,r,max;}tr[N<<2]; 6 template<typename T>void read(T&x) 7 { 8 int k=0;char c=getchar(); 9 x=0; 10 while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar(); 11 if(c==EOF)exit(0); 12 while(isdigit(c))x=x*10+c-'0',c=getchar(); 13 x=k?-x:x; 14 } 15 void bt(int x,int l,int r,int tt) 16 { 17 tr[x]=Tree{l,r,tt}; 18 if(l==r)return; 19 int mid=(l+r)>>1; 20 bt(x<<1,l,mid,tt); 21 bt(x<<1|1,mid+1,r,tt); 22 } 23 void push_up(int x) 24 { 25 tr[x].max=max(tr[x<<1].max,tr[x<<1|1].max); 26 } 27 int update(int x,int tt) 28 { 29 if (tr[x].max<tt)return -1; 30 if(tr[x].l==tr[x].r) 31 { 32 tr[x].max-=tt; 33 return tr[x].l; 34 } 35 int ans; 36 if(tr[x<<1].max>=tt) 37 ans=update(x<<1,tt); 38 else ans=update(x<<1|1,tt); 39 push_up(x); 40 return ans; 41 } 42 void work() 43 { 44 read(h);read(w);read(n); 45 if (h>n)h=n; 46 bt(1,1,h,w); 47 for(int i=1;i<=n;i++) 48 { 49 int x,ans; 50 read(x); 51 ans=update(1,x); 52 printf("%d ",ans); 53 } 54 } 55 int main() 56 { 57 #ifndef ONLINE_JUDGE 58 freopen("aa.in","r",stdin); 59 #endif 60 while(1) 61 { 62 work(); 63 } 64 }