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+;
 const int N = 1e5+;
 struct Node{
     int a, id;
     int x, y;
 }A[*];
 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[][];
 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 = ;
     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 = ;
     for(int i = ; i <= n; i++)
         for(int j = ; 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+, A++t, cmp);
     LL ans = ;
     for(int i = ; 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 ;
 }

Another Version of Inversion