逆序对

树状数组求法(权值树状数组)

(O((N + M) log M))
对于值域较大的要离散化

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 40005;
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
int n, dat[MAXN], id[MAXN], sub[MAXN], c[MAXN], size;
int lowbit(const int & x){
	return x & -x;
}
int query(int x) {
	int ans = 0;
	for( ; x; x -= lowbit(x)) ans += c[x];
	return ans;
}
void add(int x, int y) {
	for( ; x <= size; x += lowbit(x)) c[x] += y;
}
int main() {
	n = init();
	for(int i = 1; i <= n; i++) {
		dat[i] = sub[i] = init();
	}
	sort(sub + 1, sub + n + 1);
	size = unique(sub + 1, sub + n + 1) - sub - 1;
	for(int i = 1; i <= n; i++) {
		id[i] = lower_bound(sub + 1, sub + size + 1, dat[i]) - sub;
	}
	int ans = 0;
	for(int i = n; i >= 1; i--) {
		add(id[i], 1);
		ans += query(id[i] - 1);
	}
	cout << ans << endl;
	return 0;
}

归并排序求逆序对

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int MAXN = 40005;
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
int n, m, a[MAXN], b[MAXN], ans;
void merge_sort(int l, int r) {
	if(l == r) return;
	int mid = (l + r) >> 1;
	merge_sort(l, mid);
	merge_sort(mid + 1, r);
	int i = l, j = mid + 1;
	for(int k = l; k <= r; k++) {
		if(j > r || (i <= mid && a[i] < a[j])) b[k] = a[i++];
		else b[k] = a[j++], ans += mid - i + 1;
	}
	for(int i = l; i <= r; i++) a[i] = b[i];
}
int main() {
	n = init();
	for(int i = 1; i <= n; i++) a[i] = init();
	merge_sort(1, n);
	//for(int i = 1;i <= n; i++) cout << a[i] << " ";
	cout << ans << endl;
	return 0;
}