hdu 1394 Minimum Inversion Number（线段树）

参考：http://blog.sina.com.cn/s/blog_691ce2b70101ldmm.html

https://blog.csdn.net/wiking__acm/article/details/7920429

语法：https://zhidao.baidu.com/question/172532047.html

 #include <stdio.h>
 #include <algorithm>

 using namespace std;

 int a[];
 struct Node{
     int l,r,num;//num该区间已出现数的个数
 }tree[];

 void Build(int n,int x,int y){
     tree[n].l = x;
     tree[n].r = y;
     tree[n].num = ;
     if(x == y){
         return;
     }
     int mid = (x + y) / ;
     Build(*n,x,mid);
     Build(*n+,mid+,y);
 }

 void Modify(int n,int x){//在输入的相应叶节点加1
     int l = tree[n].l;
     int r = tree[n].r;
     int mid = (l + r) / ;
     if(x == l && x == r){
         tree[n].num = ;
         return;
     }
     if(x <= mid)    Modify(*n,x);
     else            Modify(*n+,x);
     tree[n].num = tree[*n].num + tree[*n+].num;
 }

 int Query(int n,int x,int y){//求大于该数且已经出现的数个数，n节点序号，x左端点，y右端点
     int l = tree[n].l;
     int r = tree[n].r;
     int mid = (l + r) / ;
     int ans = ;
     if(x == l && y == r)
         return tree[n].num;
     if(x <= mid)   ans += Query(*n,x,min(mid,y));
     if(y > mid)    ans += Query(*n+,max(mid+,x),y);
     return ans;
 }
 int main(){
     int n,sum,ans;
     int i,j;

     while(scanf("%d",&n) != EOF){
         sum = ;
         Build(,,n-);//区间端点取0~n-1即可
         for(i = ;i <= n;i++){
             scanf("%d",&a[i]);
             Modify(,a[i]);
             sum += Query(,a[i]+,n-);
         }
         ans = sum;
         for(i = ;i < n;i++){
             sum = sum + (n -  - a[i]) - a[i];//根据递推式加上接下来依次循环的序列的逆序数
             if(sum < ans)
                 ans = sum;//要最小的
         }
         printf("%d\n",ans);
     }
 }