[BZOJ4989] [Usaco2017 Feb]Why Did the Cow Cross the Road(树状数组)
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发现就是逆序对
可以树状数组求出
对于旋转操作,把一个序列最后面一个数移到开头,假设另一个序列的这个数在位置x,那么对答案的贡献 - (n - x) + (x - 1)
#include <cstdio> #include <cstring> #include <iostream> #define N 200011 #define LL long long using namespace std; int n; int a[N], b[N], pos1[N], pos2[N]; LL c[N], ans = 1ll * N * N; inline int read() { int x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1; for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘; return x * f; } inline void add(int x) { for(; x <= n; x += x & -x) c[x]++; } inline LL query(int x) { LL ret = 0; for(; x; x -= x & -x) ret += c[x]; return ret; } inline void solve() { int i, x; LL sum = 0; memset(c, 0, sizeof(c)); for(i = 1; i <= n; i++) pos1[a[i]] = i; for(i = 1; i <= n; i++) pos2[b[i]] = i; for(i = 1; i <= n; i++) { x = pos1[b[i]]; sum += (LL)i - 1 - query(x); add(x); } ans = min(ans, sum); for(i = n; i > 1; i--) { x = pos2[a[i]]; sum -= n - x; sum += x - 1; ans = min(ans, sum); } } int main() { int i; n = read(); for(i = 1; i <= n; i++) a[i] = read(); for(i = 1; i <= n; i++) b[i] = read(); solve(); swap(a, b); solve(); printf("%lld\n", ans); return 0; }
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