HDU 1394 Minimum Inversion Number 树状数组/线段树

题意：给定一个数组 a1, a2, a3, ------ an, （是一个0-n-1的排列）。可以将它转换为以下形式。每次将前m个数提取出来放至末尾。
a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)
求所有这些数列的逆序数中的最小值。
题解：树状数组求逆序数果真很给力。注意以下线段树方法的更新过程。

方法一：树状数组

#include <iostream>using namespace std;#define min(a,b) ( a < b ? a : b )#define N 5005int c[N], array[N], n;int lowbit ( int x ){return x & ( - x );}void update ( int x ){for ( int i = x; i <= n; i += lowbit(i) )c[i]++;}int cal ( int x ){int i, sum = 0;for ( i = x; i >= 1; i -= lowbit(i) )sum += c[i];return sum;}int main(){int i, ans, sum;while ( scanf("%d",&n) != EOF ){memset(c,0,sizeof(c));sum = 0;for ( i = 1; i <= n; i++ ){scanf("%d",&array[i]);array[i]++;update(array[i]);sum += i - cal(array[i]);}ans = sum;for ( i = 1; i <= n; i++ ){sum = sum - ( array[i] - 1 ) + n - array[i]; ans = min ( ans, sum );}printf("%d\n",ans);}return 0;}

方法二：线段树

#include <iostream>using namespace std;#define L(u) ( u << 1 )#define R(u) ( u << 1 | 1 )#define max(a,b) ( a > b ? a : b )#define min(a,b) ( a < b ? a : b )#define N 5005int array[N], n;struct item { int l, r, v; } node[N*3];void build ( int u, int l, int r ){node[u].l = l;node[u].r = r;node[u].v = 0;if ( l == r ) return;int mid = ( l + r ) >> 1;build ( L(u), l, mid );build ( R(u), mid+1, r );}void update ( int u, int pos ){node[u].v++;if ( node[u].l == node[u].r )return;int mid = ( node[u].l + node[u].r ) >> 1;if ( pos <= mid )update ( L(u), pos );elseupdate ( R(u), pos );}int query ( int u, int l, int r ){if ( l <= node[u].l && node[u].r <= r )return node[u].v;int mid = ( node[u].l + node[u].r ) >> 1;if ( r <= mid)return query ( L(u), l, r );else if ( l > mid )return query ( R(u), l, r );elsereturn query ( L(u), l, mid ) + query ( R(u), mid+1, r );}int main(){int i, ans, sum;while ( scanf("%d",&n) != EOF ){sum = 0;build ( 1, 1, n );for ( i = 1; i <= n; i++ ){scanf("%d",&array[i]);array[i]++;sum += query ( 1, array[i], n );update ( 1, array[i]);}ans = sum;for ( i = 1; i <= n; i++ ){sum = sum - ( array[i] - 1 ) + n - array[i]; ans = min ( ans, sum );}printf("%d\n",ans);}return 0;}