Another Version of Inversion 二维树状数组求逆序对

Another Version of Inversion

题意：只有2种走路方式，往右或者往下，求先走到一个大的数，在走到小的数的这种方式有多少。也就是说求出关于这个2维矩阵的逆序数。

题解：二维数组+逆序数就完事了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9+7;
const int N = 1e5+10;
struct Node{
    int a, id;
    int x, y;
}A[305*305];
bool cmp(Node x1, Node x2){
    if(x1.a == x2.a) return x1.id > x2.id;
    return x1.a > x2.a;
}
int lowbit(int x){
    return x&(-x);
}
int n, m;
LL tree[305][305];
void update(int x, int y){
    for(int i = x; i <= n; i+=lowbit(i))
        for(int j = y; j <= m; j += lowbit(j))
            tree[i][j]++;
}
LL query(int x, int y){
    LL ret = 0;
    for(int i = x; i; i-=lowbit(i))
        for(int j = y; j; j-=lowbit(j))
            ret += tree[i][j];
    return ret;
}
int main(){
    ///Fopen;
    scanf("%d%d", &n, &m);
    int t = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++){
            t++;
            scanf("%d", &A[t].a);
            A[t].x = i;
            A[t].y = j;
            A[t].id = t;
          //  cout <<'s' <<A[i].id << endl;
        }
    sort(A+1, A+1+t, cmp);
    LL ans = 0;
    for(int i = 1; i <= t; i++){
        //cout << A[i].id << ' ' << A[i].x << ' ' << A[i].y << endl;
        ans += query(A[i].x,A[i].y);
        update(A[i].x, A[i].y);
    }
    printf("%I64d", ans);
    return 0;
}