topcoder Single Round 624(500)

题目 

Problem Statement
	Byteland is a city with many skyscrapers, so it's a perfect venue            for BASE jumping. Danilo is an enthusiastic BASE jumper. He plans            to come to Byteland and to jump off some of its buildings.                     Danilo wants to make M jumps in Byteland. However, he has            some rules. First, he never jumps off the same building twice.            Second, all buildings he selects for his jumps must have the same            number of floors. (This is for safety reasons: It is hard to get            the timing right if each jump starts at a different height.)                     Philipe is the mayor of Byteland. He welcomes Danilo's visit as he            would like to use it as a publicity stunt. However, Philipe knows            that Danilo will only come to Byteland if there are at least M            buildings that each have the same number of floors. To ensure            that, the mayor is willing to build additional floors on some of            the skyscrapers in Byteland.                     You are given the int M and a vector <int> heights.            Each element of heights is the number of floors in one of            Byteland's skyscrapers. Compute and return the smallest number of            additional floors the mayor has to build so that there will be at            least M buildings with the same number of floors.
Definition
	Class: BuildingHeightsEasy Method: minimum Parameters: int, vector <int> Returns: int Method signature: int minimum(int M, vector <int> heights) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 256
Constraints
-	heights will contain between 1 and 50 elements, inclusive.
-	M will be between 1 and the number of elements in heights,          inclusive.
-	Each element in heights will be between 1 and 50, inclusive.
Examples
0)
	2                     {1, 2, 1, 4, 3}                     Returns: 0               Note that we already have two buildings that have the                      same number of floors. Hence, no additional floors need                      to be built.
1)
	3                     {1, 3, 5, 2, 1}                     Returns: 2               We want to have at least three buildings with the same                      number of floors. The best way to reach this goal is to                      build one floor on building #0 and one floor on building                      #4 (0-based indices). After these changes, buildings #0,                      #3, and #4 will have two floors each.
2)
	1                     {43, 19, 15}                     Returns: 0
3)
	3                     {19, 23, 9, 12}                     Returns: 15
4)
	12                     {25, 18, 38, 1, 42, 41, 14, 16, 19, 46, 42, 39, 38, 31, 43, 37, 26, 41, 33, 37, 45, 27, 19, 24, 33, 11, 22, 20, 36, 4, 4}                     Returns: 47

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解析

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代码

#include <cstring>
#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <cstring>
using namespace std;
#define MAX_HEIGHT  (50)
#define ARRAY_SIZE  (MAX_HEIGHT + 10)
#define INF         (10000000) 
 
struct BuildingHeightsEasy{
int minimum(int M, vector <int> heights){
 int min_costnum =  INF;
 memset(height_table, 0, ARRAY_SIZE * sizeof(int));
 
 for(int i = 0; i < heights.size(); i++){
   height_table[heights[i]]++;
 }
 
 for(int base_height = MAX_HEIGHT; base_height >= 1; base_height--){
   int build_num = M;
   int costnum   = 0;
   int totbuilds;
   int curr_height = base_height;
 
   while(build_num > 0 && curr_height){
     //build cost  
     totbuilds = min(build_num, height_table[curr_height]);    
     costnum   = costnum + totbuilds * (base_height - curr_height);
 
     //build num
     if(build_num <= height_table[curr_height]){
       build_num = 0;
     }else{
       build_num -= height_table[curr_height];
     }
     curr_height--;
   }
 
   if(build_num > 0){
           continue;
   }
 
   if(costnum < min_costnum){
    min_costnum = costnum;
   }
 
 }
 
 return min_costnum;
}
 
int height_table[ARRAY_SIZE]; 
 
};