標籤:codeforces 貪心 分治
題目連結:http://codeforces.com/problemset/problem/448/C
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C. Painting Fencetime limit per test1 secondmemory limit per test512 megabytesinputstandard inputoutputstandard output
Bizon the Champion isn‘t just attentive, he also is very hardworking.
Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.
Bizon the Champion bought a brush in the shop, the brush‘s width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush‘s full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.
Input
The first line contains integer n (1?≤?n?≤?5000) — the number of fence planks. The second line contains n space-separated integersa1,?a2,?...,?an (1?≤?ai?≤?109).
Output
Print a single integer — the minimum number of strokes needed to paint the whole fence.
Sample test(s)input
52 2 1 2 1
output
3
input
22 2
output
2
input
15
output
1
Note
In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.
In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.
In the third sample there is only one plank that can be painted using a single vertical stroke.
題意:
給籬笆上色,要求步驟最少,籬笆怎麼上色應該懂吧,,,刷子可以在橫著和豎著刷,不能跳著刷。
如果是豎著刷,應當是籬笆的條數,橫著刷的話,就是刷完最短木板的長度,再接著考慮沒有刷的木板中最短的。
代碼如下:
#include <iostream>#include <algorithm>using namespace std;#define INF 0x3fffffffint n, a[5017];int dfs(int sl, int sr){int MIN = INF, num = 0;if(sl > sr)return 0;for(int i = sl; i <= sr; i++){if(a[i] < MIN){MIN = a[i];}}for(int i = sl; i <= sr; i++){a[i]-=MIN;}num+=MIN;int ll = sl;for(int i = sl; i <= sr; i++){if(a[i] == 0)//如果a[i]=0,中斷處 {num+=dfs(ll,i-1);ll =i+1;}}if(ll <= sr)//最後一根不是零的情況 {num+=dfs(ll,sr);}return min(num,sr-sl+1);}int main(){int i, j;while(cin>>n){for(i = 1; i <= n; i++){cin>>a[i];}int ans = dfs(1,n);cout<<ans<<endl;}return 0;}
Codeforces Round #256 (Div. 2) C. Painting Fence(分治貪心)