CodeForces 605 A.Sorting Railway Cars(贪心)
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Description
一个1~n的排列,每次可以把其中任一个数放到序列头或尾,问最少经过多少次操作可以把该序列变成1~n
Input
第一行一整数n表示序列长度,之后n个整数a[i]表示该排列(1<=n<=1e5)
Output
输出最少操作数
Sample Input
5
4 1 2 5 3
Sample Output
2
Solution
贪心,只要固定的点最多,那么移动的点就最少,记i的位置为pos[i],那么问题转化为求pos[i] < pos[i+1] < pos[i+2] < …的最长段,扫一遍即可,时间复杂度O(n)
Code
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#include<ctime>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define maxn 111111int n,a[maxn],pos[maxn],flag[maxn];int main(){ while(~scanf("%d",&n)) { memset(flag,0,sizeof(flag)); for(int i=1;i<=n;i++)scanf("%d",&a[i]),pos[a[i]]=i; int ans=1; for(int i=1;i<=n;i++) if(!flag[i]) { int j=i,num=1; while(pos[j+1]>pos[j]) flag[j]=1,num++,j++; ans=max(ans,num); } printf("%d\n",n-ans); } return 0;}
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