HDU 4252 A Famous City

A Famous City

Time Limit: 3000ms

Memory Limit: 32768KB

This 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.

 

Input

Each 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.

 

Output

For each test case, display a single line containing the case number and the minimum possible number of buildings in the photo.

 

Sample Input

3

1 2 3

3

1 2 1

Sample Output

Case 1: 3

Case 2: 2

Hint

The possible configurations of the samples are illustrated below:

Source

Fudan Local Programming Contest 2012

 

解题：这个写法貌似最坏情况下复杂度是n*n的居然也可以过，足以见此题数据之弱。。。

 

 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 long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 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 }

View Code

 

单调栈

 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 long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 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 }

View Code