Source: PAT A1067 Sort with Swap(0, i) (25 分) Description: 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…

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…

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 判题程序 Standard 作者 CHEN, Yue 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…

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…

PAT1067. Sort with Swap(0, *) (25) 并查集 题目大意 给定一个序列, 只能进行一种操作: 任一元素与 0 交换位置, 问最少需要多少次交换. 思路 最优解就是每次 0 都和所在位置本应在的元素交换位置, 共 n - 1 次, 但是在交换中 0 可能会被交换到 0 号位置. 仔细思考一下, 其实每次换到 0 就是一个图的连通分量, 按照输入的次序和输入的值, 可以求出图共有多少个联通分量. 每个联通分量若要换回去, 需要 n - 1 次交换, 但是只能用 0 交换…

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 判题程序 Standard 作者 CHEN, Yue 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…

题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/678 5-16 Sort with Swap(0, i)   (25分) Given any permutation of the numbers {0, 1, 2,..., N-1N−1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is…

7-16 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 app…

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…

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,…

原题连接:https://pta.patest.cn/pta/test/16/exam/4/question/678 题目如下: Given any permutation of the numbers {0, 1, 2,..., N−1N-1N−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 exam…

时间限制 150 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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…

传送门:点我 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…

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 fo…

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 fo…

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 fo…

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 fo…

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 fo…

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 fo…

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…

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 fo…

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 fo…

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…

对于一个由0到N-1的序列,如果只能交换0和另一个数的位置,求多少次能够将序列变为递增序列. 输入为<N> <序列>(N和序列之间有一个空格,序列元素之间均有一个空格). 设序列存储在数组A里. 一个直接的思路是将0和0所在位置应有的元素交换,如果0到达了0号位置但是序列没有完全变成递增序列,则让0和一个没有归位的元素进行一个废交换,然后再进行后续的交换. 一个独特的思路是,将0应在位置的元素(即A[0])换到它应该在的位置(例如A[0]=3,则把A[0]和A[3]交换),不断这样…

一.技术总结 题目要求是,只能使用0,进行交换位置,然后达到按序排列,所使用的最少交换次数 输入时,用数组记录好每个数字所在的位置. 然后使用for循环,查看i当前位置是否为该数字,核心是等待0回到自己的位置上,如果没有一直与其他的数字进行交换.如果这个过程中,i回到自己位置上就不用调换,如果不是那么直接与0位置进行调换,直到所有位置都到指定位置上面. 二.参考代码 #include<iostream> #include<algorithm> using namespace std…

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…

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…

2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分子. // 1063 Set Similarity #include <set> #include <map> #include <cstdio> #include <iostream> #include <algorithm> using name…