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:  題意：利用題目中給定的規律尋找到最少建筑的數量。弄懂題意很重要......數據范圍不大，可以暴力試一試。建筑的最大可能是總數減去0高度的，則答案就是最大可能減去重復的建筑。從開始遍歷，如果當前建筑低于前一個建筑，則遍歷位于當前建筑前面的建筑，看是否有與其相等的。若當前建筑高于前一個建筑，說明當前建筑是單獨的，繼續遍歷下一個建筑。具體看代碼。AC代碼：#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])  建筑重復的情況                    {                          ans++;                          break;                      }                  }              }          }          printf("Case %d: %d/n",cas++,n-ans); ans包括 0高度建筑與重復建筑    }        return 0;  }