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UVa 1611 Crane (构造+贪心)
题意:给定一个序列,让你经过不超过9的6次方次操作,变成一个有序的,操作只有在一个连续区间,交换前一半和后一半。
析:这是一个构造题,我们可以对第 i 个位置找 i 在哪,假设 i 在pos 位置,那么如果 (pos-i)*2+i-1 <= n,那么可以操作一次换过来,
如果不行再换一种,如果他们之间元素是偶数,那么交换 i - pos,如果是奇数,交换 i - pos+1,然后再经过一次就可以换到指定位置。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e5 + 5;const int mod = 1e9 + 7;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}int pos[maxn], a[maxn];void Swap(int l, int r){ int mid = (r-l+1)/2; for(int i = 0; i < mid; ++i){ swap(pos[a[l]], pos[a[l+mid]]); swap(a[l], a[l+mid]); ++l; }}vector<P> ans;int main(){ int T; cin >> T; while(T--){ scanf("%d", &n); for(int i = 1; i <= n; ++i){ scanf("%d", a+i); pos[a[i]] = i; } ans.clear(); for(int i = 1; i <= n; ++i){ int id = pos[i]; if(i == a[i]) continue; if((id-i)*2+i-1 <= n){ Swap(i, (id-i)*2+i-1); ans.push_back(P(i, (id-i)*2+i-1)); } else if((id-i) & 1){ Swap(i, id); ans.push_back(P(i, id)); } else if(id == n){ Swap(n-1, n); ans.push_back(P(n-1, n)); } else{ Swap(i, id+1); ans.push_back(P(i, id+1)); } --i; } printf("%d\n", ans.size()); for(int i = 0; i < ans.size(); ++i) printf("%d %d\n", ans[i].first, ans[i].second); } return 0;}
UVa 1611 Crane (构造+贪心)
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