PAT|1067. Sort with Swap(0,*)
2016-09-06 15:53
411 查看
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 following way:
Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4)=> {0, 1, 2, 3, 4}
Now you are asked to find the minimum number of swaps need to sort the
given permutation of the first N nonnegative integers.
Input Specification:
Each input file contains one test case, which gives a positive N
(<=105) followed by a permutation sequence of {0, 1, …, N-1}. All
the numbers in a line are separated by a space.
Output Specification:
For each case, simply print in a line the minimum number of swaps need
to sort the given permutation.
Sample Input: 10 3 5 7 2 6 4 9 0 8 1 Sample Output: 9
并查集
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 following way:
Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4)=> {0, 1, 2, 3, 4}
Now you are asked to find the minimum number of swaps need to sort the
given permutation of the first N nonnegative integers.
Input Specification:
Each input file contains one test case, which gives a positive N
(<=105) followed by a permutation sequence of {0, 1, …, N-1}. All
the numbers in a line are separated by a space.
Output Specification:
For each case, simply print in a line the minimum number of swaps need
to sort the given permutation.
Sample Input: 10 3 5 7 2 6 4 9 0 8 1 Sample Output: 9
并查集
#include<iostream> #include<stdio.h> #include<vector> using namespace std; vector<int>s; int n; int findRoot(int x) { if (s[x] < 0) return x; else return findRoot(s[x]); } void Uion(int a,int b) { int ra = findRoot(a); int rb = findRoot(b); if (ra != rb) { if (s[ra] <= s[rb]) { s[ra] += s[rb]; s[rb] = ra; } else { s[rb] += s[ra]; s[ra] = rb; } } } int main() { while (scanf("%d",&n)!=EOF) { s.clear(); s.resize(n); for (int i = 0; i < n; i++) { s[i] = -1; } for (int i = 0; i < n; i++) { int emp; scanf("%d", &emp); Uion(i, emp); } int k = 0; int x = 0; for (int i = 0; i < n; i++) { if (s[i] < -1) k++; if (s[i] == -1) x++; } int count; if (s[0] != -1) count = (n - x) + k - 2; else count = (n - x) + k; printf("%d\n", count); } return 0; }
相关文章推荐
- 1611:The Suspects
- 作业四,1001
- 作业四1002
- 作业四1003
- 练习四1005
- 最小权值和
- HDU-1213-How Many Tables
- Longest Consecutive Sequence,Distinct Subsequences,Interleaving String,Scramble String
- 并查集_POJ 1182_食物链
- SARS病毒传染 并查集
- HDU 1213
- CSU1307 并查集+SPFA
- 并查集
- BestWiring——Kruskal算法&并查集
- 1611:The Suspects
- 并查集示例1
- 并查集(union-find)学习报告
- poj3728
- HDU-1233 还是畅通工程(最小生成树&并查集)
- Simon-【深入理解数据结构】有根树的不同实现① —— 并查集