HDU4252：A Famous City

2019-11-08 02:15:37

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PRoblem DescriptionAfter Mr. B arrived in Warsaw, he was shocked by the skyscrapers and took several photos. But now when he looks at these photos, he finds in surprise that he isn't able to point out even the number of buildings in it. So he decides to work it out as follows:- divide the photo into n vertical pieces from left to right. The buildings in the photo can be treated as rectangles, the lower edge of which is the horizon. One building may span several consecutive pieces, but each piece can only contain one visible building, or no buildings at all.- measure the height of each building in that piece.- write a program to calculate the minimum number of buildings.Mr. B has finished the first two steps, the last comes to you. InputEach test case starts with a line containing an integer n (1 <= n <= 100,000). Following this is a line containing n integers - the height of building in each piece respectively. Note that zero height means there are no buildings in this piece at all. All the input numbers will be nonnegative and less than 1,000,000,000. OutputFor each test case, display a single line containing the case number and the minimum possible number of buildings in the photo. Sample Input

31 2 331 2 1 Sample OutputCase 1: 3Case 2: 2HintThe possible configurations of the samples are illustrated below:  题意：利用题目中给定的规律寻找到最少建筑的数量。弄懂题意很重要......数据范围不大，可以暴力试一试。建筑的最大可能是总数减去0高度的，则答案就是最大可能减去重复的建筑。从开始遍历，如果当前建筑低于前一个建筑，则遍历位于当前建筑前面的建筑，看是否有与其相等的。若当前建筑高于前一个建筑，说明当前建筑是单独的，继续遍历下一个建筑。具体看代码。AC代码：#include <stdio.h>  #include <string.h>  #include <algorithm>  using namespace std;    int a[1000005],x,len;    int main()  {      int n,i,cas = 1,ans,j;      while(~scanf("%d",&n))      {          ans = 0;          for(i = 0; i<n; i++)              scanf("%d",&a[i]);          if(a[0]==0)              ans++;          for(i = 1; i<n; i++)          {              if(a[i]==0)                  ans++;  0高度建筑            else              {                  for(j = i-1; j>=0; j--)                  {                      if(a[i]>a[j])                          break;                      else if(a[i]==a[j])  建筑重复的情况                    {                          ans++;                          break;                      }                  }              }          }          printf("Case %d: %d/n",cas++,n-ans); ans包括 0高度建筑与重复建筑    }        return 0;  }