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Description
Some of Farmer John‘s N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows‘ heads.
Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.
Consider this example:
=
= =
= - = Cows facing right -->
= = =
= - = = =
= = = = = =
1 2 3 4 5 6
Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow‘s hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow‘s hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!
Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Input
Output
Sample Input
6
10
3
7
4
12
2
Sample Output
5
题意： 有n头牛从左到右排成一行，从左到右编号为1~n，并且牛的头是朝向右的，即每头牛只可以看到他们右边的牛。
现在给出n头牛的身高，设c[i]表示第i头牛可以看到多少头牛的头顶。
求所有c[i]的和。
简单DP
设dp[i]表示第i头牛在看的时候是被第dp[i]头牛挡住视线的，即i+1~dp[i]-1这个区间的牛都可以被第i头牛看到。
所以第i头牛看到的数目为：dp[i]-i-1
这道题也可以用栈来做。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 5 const int maxn=80000+5; 6 const long long inf=0x3f3f3f3f; 7 8 #define LL long long 9 10 LL h[maxn]; 11 LL dp[maxn]; 12 13 int main() 14 { 15 int n; 16 while(~scanf("%d",&n)) 17 { 18 for(int i=1;i<=n;i++) 19 { 20 scanf("%lld",&h[i]); 21 } 22 n++; 23 h[0]=h[n]=inf; 24 25 for(int i=n-1;i>0;i--) 26 { 27 int tmp=i+1; 28 dp[i]=i+1; 29 while(h[i]>h[tmp]) 30 { 31 dp[i]=dp[tmp]; 32 tmp=dp[tmp]; 33 } 34 } 35 LL ans=0; 36 for(int i=1;i<n;i++) 37 { 38 ans+=(dp[i]-i-1); 39 } 40 printf("%lld\n",ans); 41 } 42 return 0; 43 }
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原文地址：http://www.cnblogs.com/-maybe/p/4539561.html
