HDU 1394Minimum Inversion Number

题意：很水的用线段树求逆序对的题，时间复杂度在nlogn适合初学者，数据较小不用离散化，直接敲即可；

#include<algorithm>
#include<iostream>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<queue>
#include<cstring>
#include<cstdio>
#define N 10005
#define INF 0x3f3f3f3f
using namespace std;
typedef struct node{
	int x;int y;int date;
}node;
node a[5005*4];
void built(int root,int first,int end){
	if(first==end){
		a[root].x=first;a[root].y=end;a[root].date=0;
		return ;
	}
	int mid=(first+end)/2;
	built(root*2,first,mid);
	built(root*2+1,mid+1,end);
	a[root].x=a[root*2].x;a[root].y=a[root*2+1].y;a[root].date=0;
}
void U(int root,int first,int end,int e){
	if(first==end){
		a[root].date=1;
		return ;
	}
	int mid=(first+end)/2;
	if(e<=mid)  U(root*2,first,mid,e);
	else U(root*2+1,mid+1,end,e);
	a[root].date=a[root*2].date+a[root*2+1].date;
}
int sum=0;
void Q(int root,int first,int end,int l,int r){
	if(first>=l&&end<=r){
		sum+=a[root].date;
		return ;
	}
	int mid=(first+end)/2;
	if(l<=mid)  Q(root*2,first,mid,l,r);
	if(r>mid)  Q(root*2+1,mid+1,end,l,r);
}
int b[5005];
int c[5005];
int main(){
	int n;
	while(scanf("%d",&n)==1){
		for(int i=1;i<=n;i++){
			scanf("%d",&b[i]);
		}
		int ans=0;
		built(1,0,n-1);
		for(int i=1;i<=n;i++){
			sum=0;
			if(b[i]==0){
				U(1,0,n-1,0);
			}
			else{
				Q(1,0,n-1,0,b[i]-1);
				ans+=(b[i]-sum);
				U(1,0,n-1,b[i]);
			}
		}
		c[1]=ans;
		//cout<<ans<<endl; 
		for(int i=2;i<=n;i++){
			c[i]=c[i-1]-b[i-1]-b[i-1]+n-1;
		}
		int p=INF;
		for(int i=1;i<=n;i++){
			p=min(p,c[i]);
		}
		printf("%d
",p);
} 
  return 0;
}