深入理解线段树运作流程(2795)

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17263    Accepted Submission(s): 7294

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

乍一看h范围很大，但是n范围最多20W，所以，多出来的很多行木板是多余的。通过这道题要深刻理解线段树的运作流程：原题要求有相同满足条件时，优先选择靠前的，所以我们在遍历的时候一定要先从左子树开始考虑，tree[1]保存着整个区间的最大值，所以如果最大值也比要插入的小，就可以直接输出-1了。

/*------------------Header Files------------------*/#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <cstdlib>#include <ctype.h>#include <cmath>#include <stack>#include <queue>#include <deque>#include <map>#include <vector>#include <limits.h>using namespace std;/*------------------Definitions-------------------*/#define LL long long#define PI acos(-1.0)#define INF 0x3F3F3F3F#define MOD 1E9+7#define MAX 500050#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r/*---------------------Work-----------------------*/int h,w,n,tree[200050<<2];void pushup(int rt){tree[rt]=max(tree[rt<<1],tree[rt<<1|1]);}void build(int rt,int l,int r){tree[rt]=w;if(l==r) return ;int m=(l+r)>>1;build(lson);build(rson);}int query(int rt,int l,int r,int x){if(l==r) //肯定要遍历到一个叶子结点为止的{tree[rt]-=x;return l;}int m=(l+r)>>1,ans;if(tree[rt<<1]>=x) ans=query(lson,x); //左子树优先考虑else ans=query(rson,x);pushup(rt);return ans;}void work(){while(scanf("%d%d%d",&h,&w,&n)==3){if(h>n) h=n; //这一步很关键build(1,1,h);int ans;while(n--){scanf("%d",&ans);if(tree[1]<ans) printf("-1\n");else printf("%d\n",query(1,1,h,ans));}}}/*------------------Main Function------------------*/int main(){//freopen("test.txt","r",stdin);//freopen("cowtour.out","w",stdout);//freopen("cowtour.in","r",stdin);work();return 0;}