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

Posted 2020-11-09 Schenker

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：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 4 #define LL long long
 5 #define ULL unsigned LL
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 #define lson l,m,rt<<1
10 #define rson m+1,r,rt<<1|1
11 #define max3(a,b,c) max(a,max(b,c))
12 #define min3(a,b,c) min(a,min(b,c))
13 typedef pair<int,int> pll;
14 const int INF = 0x3f3f3f3f;
15 const LL mod = 1e9+7;
16 const int N = 1e5+10;
17 struct Node{
18     int a, id;
19     int x, y;
20 }A[305*305];
21 bool cmp(Node x1, Node x2){
22     if(x1.a == x2.a) return x1.id > x2.id;
23     return x1.a > x2.a;
24 }
25 int lowbit(int x){
26     return x&(-x);
27 }
28 int n, m;
29 LL tree[305][305];
30 void update(int x, int y){
31     for(int i = x; i <= n; i+=lowbit(i))
32         for(int j = y; j <= m; j += lowbit(j))
33             tree[i][j]++;
34 }
35 LL query(int x, int y){
36     LL ret = 0;
37     for(int i = x; i; i-=lowbit(i))
38         for(int j = y; j; j-=lowbit(j))
39             ret += tree[i][j];
40     return ret;
41 }
42 int main(){
43     ///Fopen;
44     scanf("%d%d", &n, &m);
45     int t = 0;
46     for(int i = 1; i <= n; i++)
47         for(int j = 1; j <= m; j++){
48             t++;
49             scanf("%d", &A[t].a);
50             A[t].x = i;
51             A[t].y = j;
52             A[t].id = t;
53           //  cout <<‘s‘ <<A[i].id << endl;
54         }
55     sort(A+1, A+1+t, cmp);
56     LL ans = 0;
57     for(int i = 1; i <= t; i++){
58         //cout << A[i].id << ‘ ‘ << A[i].x << ‘ ‘ << A[i].y << endl;
59         ans += query(A[i].x,A[i].y);
60         update(A[i].x, A[i].y);
61     }
62     printf("%I64d", ans);
63     return 0;
64 }

Another Version of Inversion