[HDU] 2795 Billboard [线段树区间求最值]
Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11861 Accepted Submission(s): 5223
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.
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.
题意:h*w的木板,放进一些1*L的物品,求每次放空间能容纳且最上边的位子
思路:每次找到最大值的位子,然后减去L
线段树功能:query:区间求最大值的位子
#include<cstdio>
#include<string.h>
#include<algorithm> #define clr(x,y) memset(x,y,sizeof(x))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1 const int N=2e5+;
using namespace std; int h,w,n,MAX[N<<]; void PushUp(int rt)
{
MAX[rt]=max(MAX[rt<<],MAX[rt<<|]);
} void build(int l,int r,int rt)
{
int m; MAX[rt]=w;
if(l==r) {
return;
} m=(l+r)>>;
build(lson);
build(rson);
} int query(int x,int l,int r,int rt)
{
int m,ret;
if(l==r) {
MAX[rt]-=x;
return l;
}
m=(l+r)>>;
ret=(MAX[rt<<]>=x) ? query(x,lson) : query(x,rson);
PushUp(rt);
return ret;
} int main()
{
int x;
while(~scanf("%d%d%d",&h,&w,&n)) {
if(h>n) h=n;
build(,h,);
while(n--) {
scanf("%d",&x);
if(MAX[]<x) puts("-1");
else printf("%d\n",query(x,,h,)); } } return ;
}
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