P4378[USACO18OPEN]Out of Sorts S

题目

https://www.luogu.org/problemnew/show/P4378

• Problem Description
  Keeping an eye on long term career possibilities beyond the farm, Bessie the cow has started learning algorithms from various on-line coding websites.

  Her favorite algorithm thus far is “bubble sort”. Here is Bessie’s implementation, in cow-code, for sorting an array A of length N
  
   sorted = false
    while (not sorted):
     sorted = true
     moo
     for i = 0 to N-2:
      if A[i+1] < A[i]:
      swap A[i], A[i+1]
      sorted = false

  Apparently, the “moo” command in cow-code does nothing more than print out “moo”. Strangely, Bessie seems to insist on including it at various points in her code.

  Given an input array, please predict how many times “moo” will be printed by Bessie’s code.

• Input
  The first line of input contains N (1≤N≤100,000). The next N lines describe A[0]…A[N−1], each being an integer in the range 0…10⁹. Input elements are not guaranteed to be distinct.

• Output
  Print the number of times “moo” is printed.

• input
  5
  1
  5
  3
  8
  2

• output
  4

题意

求题目所给冒泡算法的循环次数

思路

每次冒泡都将最大的未排序好的数放到最后，而比较小的数会被动的往前移动一位，所以答案就是被搬到前面的数中离初始位置最远的距离+1

代码

#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000+1000;

struct node_ {
    public:
        bool friend operator<(node_ a,node_ b) {
        	if(a.val == b.val) return a.p<b.p;
            return a.val<b.val;
        }
        int val;
        int p;
} node[100010];

int main() {
    int N;
    scanf("%d",&N);

    for(int i=1; i<=N; ++i) {
        scanf("%d",&node[i].val);
        node[i].p=i;
    }
    sort(node+1,node+1+N);
    
    int maxp=0;
    for(int i=1;i<=N;++i){
    	if(i<node[i].p)
			maxp=max(maxp,node[i].p-i);
	}
	cout << maxp+1 <<endl;
    return 0;
}