POJ 3250 Bad Hair Day 简单DP 好题
2015-05-29 22:48
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DescriptionSome 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:
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.InputLine 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
OutputLine 1: A single integer that is the sum of c1 through cN.
Sample Input
Sample Output
= = = = - = 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.InputLine 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
OutputLine 1: A single integer that is the sum of c1 through cN.
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。
#include<cstdio> #include<cstring> #include<algorithm> const int maxn=80000+5; const long long inf=0x3f3f3f3f; #define LL long long LL h[maxn]; LL dp[maxn]; int main() { int n; while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) { scanf("%lld",&h[i]); } n++; h[0]=h =inf; for(int i=n-1;i>0;i--) { int tmp=i+1; dp[i]=i+1; while(h[i]>h[tmp]) { dp[i]=dp[tmp]; tmp=dp[tmp]; } } LL ans=0; for(int i=1;i<n;i++) { ans+=(dp[i]-i-1); } printf("%lld\n",ans); } return 0; }188ms
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