这道题看了半天没看出什么规律, 然后看到别人的博客, 结论是当n为奇数且逆序数为奇数的时候
无解, 否则有解。但是没有给出证明, 在网上也找到详细的证明……我也不知道是为什么……
求逆序对有两种方法, 树状数组和归并排序, 当然这道题数据很小可以直接暴力, 我三种都写了。
暴力
#include<cstdio> #define REP(i, a, b) for(int i = (a); i < (b); i++) using namespace std; const int MAXN = 512; int a[MAXN]; int main() { int T, n; scanf("%d", &T); while(T--) { int cnt = 0; scanf("%d", &n); REP(i, 0, n) { scanf("%d", &a[i]); REP(j, 0, i) if(a[j] > a[i]) cnt++; } printf("%s\n", (n & 1 && cnt & 1) ? "impossible" : "possible"); } return 0; }
归并排序
#include<cstdio> #define REP(i, a, b) for(int i = (a); i < (b); i++) using namespace std; const int MAXN = 512; int a[MAXN], t[MAXN], cnt; void merge_sort(int l, int r) { if(l + 1 >= r) return; int m = (l + r) >> 1; merge_sort(l, m); merge_sort(m, r); int p = l, q = m, i = l; while(p < m || q < r) { if(q >= r || p < m && a[p] < a[q]) t[i++] = a[p++]; else t[i++] = a[q++], cnt += m - p; } REP(i, l, r) a[i] = t[i]; } int main() { int T, n; scanf("%d", &T); while(T--) { cnt = 0; scanf("%d", &n); REP(i, 0, n) scanf("%d", &a[i]); merge_sort(0, n); printf("%s\n", (n & 1 && cnt & 1) ? "impossible" : "possible"); } return 0; }
树状数组
#include<cstdio> #include<cstring> #define REP(i, a, b) for(int i = (a); i < (b); i++) using namespace std; const int MAXN = 512; int a[MAXN], f[MAXN], n; int lowbit(int x) { return x & (-x); } void add(int x) { while(x <= n) { f[x]++; x += lowbit(x); } } int sum(int x) { int ret = 0; while(x) { ret += f[x]; x -= lowbit(x); } return ret; } int main() { int T; scanf("%d", &T); while(T--) { int cnt = 0; memset(f, 0, sizeof(f)); scanf("%d", &n); REP(i, 0, n) { scanf("%d", &a[i]); cnt += sum(n) - sum(a[i]-1); add(a[i]); } printf("%s\n", (n & 1 && cnt & 1) ? "impossible" : "possible"); } return 0; }