主題鏈接：

啊哈哈，點我點我

題意：

給一個僅僅含有1。2的序列，如何變換n次使序列成為一個非遞減的序列，而且使n最小。

思路：

這道題的數據范圍是50000，則肯定承受不了n方的復雜度。所以 僅僅能寫O(n)的算法，甚至更小，所以當時想二分，可是不知道怎么寫，忽然想到能夠枚舉每個位置，把每個位置都當做一個分界點。然后求前半部有多少個2。后半段有多少個1，最后和所有是1和2進行比較，這個問題便得到了解決。

題目：

Dining Cows

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?7237	?	Accepted:?3078

Description

The cows are so very silly about their dinner partners. They have organized themselves into two groups (conveniently numbered 1 and 2) that insist upon dining together in order, with group 1 at the beginning of the line and group 2 at the end. The trouble starts when they line up at the barn to enter the feeding area.

Each cow?i?carries with her a small card upon which is engraved?D_i?(1 ≤?D_i?≤ 2) indicating her dining group membership. The entire set of?N?(1 ≤?N?≤ 30,000) cows has lined up for dinner but it's easy for anyone to see that they are not grouped by their dinner-partner cards.

FJ's job is not so difficult. He just walks down the line of cows changing their dinner partner assignment by marking out the old number and writing in a new one. By doing so, he creates groups of cows like 112222 or 111122 where the cows' dining groups are sorted in ascending order by their dinner cards. Rarely he might change cards so that only one group of cows is left (e.g., 1111 or 222).

FJ is just as lazy as the next fellow. He's curious: what is the absolute minimum number of cards he must change to create a proper grouping of dining partners? He must only change card numbers and must not rearrange the cows standing in line.

Input

* Line 1: A single integer:?N
* Lines 2..N+1: Line?i+1 describes cow?i's dining preference with a single integer:?D_i

Output

* Line 1: A single integer that is the minimum number of cards Farmer John must change to assign the cows to eating groups as described.

Sample Input

7 2 1 1 1 2 2 1

Sample Output

2

Source

USACO 2008 February Bronze

代碼

#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #define INF 0x3f3f3f3f using namespace std; const int maxn=30000+10;int sum1[maxn],sum2[maxn];int main() {int n,cal1,cal2,tmp,ans;while(~scanf("%d",&n)){cal1=cal2=0;ans=INF;memset(sum1,0,sizeof(sum1));memset(sum2,0,sizeof(sum2));for(int i=1;i<=n;i++){scanf("%d",&tmp);if(tmp==1)sum1[i]=++cal1;elsesum1[i]=cal1;if(tmp==2)sum2[i]=++cal2;elsesum2[i]=cal2;}for(int i=1;i<n;i++)ans=min(ans,sum2[i]+(sum1[n]-sum1[i]));ans=min(ans,sum1[n]);ans=min(ans,sum2[n]);printf("%d\n",ans);}return 0; }

版權聲明：本文博客原創文章，博客，未經同意，不得轉載。

轉載于:https://www.cnblogs.com/blfshiye/p/4714796.html

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