hdu 5497 Inversion(高效)

解题思路

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e5 + 5;
typedef long long ll;
#define lowbit(x) ((x)&(-x))

struct Fenw {
int n, f[maxn];
void init(int n) {
this->n = n;
memset(f, 0, sizeof(f));
}

void add(int x, int v) {
while (x <= n) {
f[x] += v;
x += lowbit(x);
}
}

int sum(int x) {
int ret = 0;
while (x) {
ret += f[x];
x -= lowbit(x);
}
return ret;
}

int query (int l, int r) { return sum(r) - sum(l-1); }
}P, S;

int N, M, L[maxn], R[maxn], A[maxn];
ll G;

void init () {
G = 0;
scanf("%d%d", &N, &M);
for (int i = 0; i < N; i++) scanf("%d", &A[i]);

P.init(N+1);
for (int i = 0; i < N; i++) {
L[i] = P.query(A[i] + 1, N + 1);
P.add(A[i], 1);
G = G + L[i];
}
P.init(N+1);
for (int i = N-1; i >= 0; i--) {
R[i] = P.query(1, A[i] - 1);
P.add(A[i], 1);
}
}

ll solve () {
P.init(N+1);
S.init(N+1);

ll sum = 0;
for (int i = 0; i < M; i++) {
sum += R[i];
S.add(A[i], 1);
}

ll ret = sum;
for (int i = M; i < N; i++) {
sum += R[i] - R[i-M] - L[i-M];
S.add(A[i-M], -1);
S.add(A[i], 1);
sum += S.query(1, A[i-M]-1) + P.query(A[i]+1, N+1);
P.add(A[i-M], 1);
ret = max(ret, sum);
}

//printf("%lld %lld\n", ret, G);
return G - ret;
}

int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
init ();
printf("%lld\n", solve());
}
return 0;
}