Hdus 4252 a famous city

A famous citytime limit: 3000 msmemory limit: 32768 kbthis problem will be judged on HDU. Original ID: 4252
64-bit integer Io format: % i64d Java class name: Main after 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 output

Case 1: 3Case 2: 2

HintThe possible deployments of the samples are using strated below: Sourcefudan local programming contest 2012: It seems that the complexity is N * n in the worst case, which is enough to see the weak data in this question...

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 100010;18 int d[maxn],n,m;19 int main(){20     int cs = 1;21     while(~scanf("%d",&n)){22         for(int i = 0; i < n; ++i){23             scanf("%d",d+i);24         }25         m = n;26         if(d[0] == 0) --m;27         for(int i = 1; i < n; ++i){28             if(d[i] == 0) m--;29             else{30                 for(int j = i-1; j >= 0; --j){31                     if(d[j] < d[i]) break;32                     if(d[j] == d[i]){33                         --m;34                         break;35                     }36                 }37             }38         }39         printf("Case %d: %d\n",cs++,m);40     }41     return 0;42 }

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Monotonous Stack

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 stack<int>stk;18 int main(){19     int cs = 1,tmp,n;20     while(~scanf("%d",&n)){21         while(!stk.empty()) stk.pop();22         int ans = 0;23         for(int i = 0; i < n; ++i){24             scanf("%d",&tmp);25             while(!stk.empty() && stk.top() > tmp){26                 ans++;27                 stk.pop();28             }29             if(tmp && (stk.empty() || tmp > stk.top())) stk.push(tmp);30         }31         printf("Case %d: %d\n",cs++,ans + stk.size());32     }33     return 0;34 }

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Hdus 4252 a famous city