PAT 1067 Sort with Swap[难]】的更多相关文章

1067 Sort with Swap(0,*) (25)(25 分) Given any permutation of the numbers {0, 1, 2,..., N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may…
1067. Sort with Swap(0,*) (25)   Given any permutation of the numbers {0, 1, 2,..., N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may ap…
题目信息 1067. Sort with Swap(0,*) (25) 时间限制150 ms 内存限制65536 kB 代码长度限制16000 B Given any permutation of the numbers {0, 1, 2,-, N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For exa…
1067 Sort with Swap(0, i) (25 分)   Given any permutation of the numbers {0, 1, 2,..., N−1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may…
1067 Sort with Swap(0, i) (25 分) Given any permutation of the numbers {0, 1, 2,..., N−1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may ap…
题目链接:1067 Sort with Swap(0, i) (25 分) 题意 给定长度为 \(n\) 的排列,如果每次只能把某个数和第 \(0\) 个数交换,那么要使排列是升序的最少需要交换几次. 思路 贪心 由于是排列,所以排序后第 \(i\) 个位置上的数就是 \(i\).所以当 \(a[0] \neq 0\) 时,把 \(a[0]\) 位置上的元素交换到相应位置.如果 \(a[0] = 0\),就找到第一个不在正确位置上的数,把它与第 \(0\) 个数交换,那么下一次又是第一种情况了,…
题目 Given any permutation of the numbers {0, 1, 2,-, N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the f…
只对没有归位的数进行交换. 分两种情况: 如果0在最前面,那么随便拿一个没有归位的数和0交换位置. 如果0不在最前面,那么必然可以归位一个数字,将那个数字归位. 这样模拟一下即可. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include…
题意: 输入一个正整数N(<=100000),接着输入N个正整数(0~N-1的排列).每次操作可以将0和另一个数的位置进行交换,输出最少操作次数使得排列为升序. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ],b[]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(…
贪心算法 次数最少的方法,即:1.每次都将0与应该放置在0位置的数字交换即可.2.如果0处在自己位置上,那么随便与一个不处在自己位置上的数交换,重复上一步即可.拿样例举例:   0 1 2 3 4 5 6 7 8 90:3 5 7 2 6 4 9 0 8 1 1:3 5 0 2 6 4 9 7 8 12:3 5 2 0 6 4 9 7 8 13:0 5 2 3 6 4 9 7 8 1 (如果0处在自己位置上,那么找一个不在自己位置上的数val与之交换)4:5 0 2 3 6 4 9 7 8 15…