hdu 2795 Billboard

Billboard

Time Limit: ²⁰⁰⁰⁰⁄₈₀₀₀ MS (Java/Others) Memory Limit: ³²⁷⁶⁸⁄₃₂₇₆₈ K (Java/Others) Total Submission(s): 5912 Accepted Submission(s): 2682

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 5 2 4 3 3 3

Sample Output 1 2 1 3 -1

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
const int maxn = 222222;
int MAX[maxn<<2];
int h, w, n;
void PushUP(int rt){
  MAX[rt] = max(MAX[rt<<1], MAX[rt<<1|1]);
}
void build(int l, int r, int rt){
  MAX[rt] = w;
  if (l == r) return;
  int m = (l + r) >> 1; build(lson); build(rson);
}
int query(int x, int l, int r, int rt){
  if (l == r){MAX[rt] -= x; return l;}
  int m = (l + r) >> 1;
  int ret = (MAX[rt<<1] >= x) ? query(x, lson) : query(x, rson);
  PushUP(rt);
  return ret;
}
int main(void){
#ifndef ONLINE_JUDGE
  freopen("hdu2795.in", "r", stdin);
#endif
  while(~scanf("%d%d%d", &h, &w, &n)){
    if (h > n) h = n;
    build(1, h, 1);
    while (n--){
      int x; scanf("%d", &x);
      if (MAX[1] < x) printf("-1\n");
      else {printf("%d\n", query(x, 1, h, 1));}
    }
  }
  return 0;
}

树状数组，这个题目要注意，更新的时候，要先找到要更新的点，在query中实现……

也可以在写一个update，发现其实和query的代码很相似，如下：

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
const int maxn = 222222;
int MAX[maxn<<2];
int h, w, n;
void PushUP(int rt){
  MAX[rt] = max(MAX[rt<<1], MAX[rt<<1|1]);
}
void build(int l, int r, int rt){
  MAX[rt] = w;
  if (l == r) return;
  int m = (l + r) >> 1; build(lson); build(rson);
}
void update(int x, int l, int r, int rt){
  if (l == r) {MAX[rt] -= x; return;}
  int m = (l + r) >> 1;
  if (x <= MAX[rt<<1]) update(x, lson); else update(x, rson);
  PushUP(rt);
}
int query(int x, int l, int r, int rt){
  if (l == r){return l;}
  int m = (l + r) >> 1;
  int ret = (MAX[rt<<1] >= x) ? query(x, lson) : query(x, rson);
  return ret;
}
int main(void){
#ifndef ONLINE_JUDGE
  freopen("hdu2795.in", "r", stdin);
#endif
  while(~scanf("%d%d%d", &h, &w, &n)){
    if (h > n) h = n;
    build(1, h, 1);
    while (n--){
      int x; scanf("%d", &x);
      if (MAX[1] < x) printf("-1\n");
      else {printf("%d\n", query(x, 1, h, 1));update(x, 1, h, 1);}
    }
  }
  return 0;
}

理解思想就可以了。