HDU 1394 Minimum Inversion Number tree array & amp; line segment tree

The question gives you a series, and each time the last number is raised to the beginning, until the original first number reaches the last one, each operation will generate a new series, this sequence has a value in the descending order and asks the minimum value in the descending order.

The number of reverse orders is best thought of as a tree array, which is very fast. Pay attention to the impact of bringing the last number up on the number of reverse orders,

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              #define ll long long#define LL __int64#define eps 1e-8#define inf 0xfffffff//const LL INF = 1LL<<61;using namespace std;//vector
              
                > G;//typedef pair
               
                 P;//vector
                
                  > ::iterator iter;////map
                 
                  mp;//map
                  
                   ::iterator p;int n;int c[10000 + 5];int num[10000 + 5];void init() {memset(c,0,sizeof(c));memset(num,0,sizeof(num));}int lowbit(int x) {return x&(-x);}void add(int i,int val) {while(i <= n) {c[i] += val;i += lowbit(i);}}int get_sum(int i) {int sum = 0;while(i > 0) {sum += c[i];i -= lowbit(i);}return sum;}int main() {while(scanf("%d",&n) == 1) {init();int ans = 0;for(int i=1;i<=n;i++) {scanf("%d",&num[i]);num[i]++;add(num[i],1);ans += (i - get_sum(num[i]));}int minn = ans;for(int i=n;i>1;i--) {ans = ans + num[i] + num[i] - n - 1;minn = min(ans,minn);}printf("%d\n",minn);}return 0;}
                  
                 
                
               
              
             
            
           
          
         
       
      
     
    
   
  
 

Line Segment tree:

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              #define ll long long#define LL __int64#define eps 1e-8#define inf 0xfffffff//const LL INF = 1LL<<61;using namespace std;//vector
              
                > G;//typedef pair
               
                 P;//vector
                
                  > ::iterator iter;////map
                 
                  mp;//map
                  
                   ::iterator p;const int N = 10000 + 5;int num[N];typedef struct Node { int a; int l,r;};Node tree[N * 4];void init() { memset(tree,0,sizeof(tree)); memset(num,0,sizeof(num));}void cal(int id) { tree[id].a = min(tree[id<<1].a,tree[id<<1|1].a);}void build(int l,int r,int id) { tree[id].l = l; tree[id].r = r; tree[id].a = 0; if(l == r) return ; int mid = (l + r)/2; build(l,mid,id<<1); build(mid+1,r,id<<1|1);}void updata(int w,int id) { if(tree[id]. l == w && tree[id].r == w) { tree[id].a = 1;return; } int mid = (tree[id].l + tree[id].r)/2; if(w <= mid) updata(w,id<<1); else updata(w,id<<1|1); tree[id].a = tree[id<<1].a + tree[id<<1|1].a;}int query(int l,int r,int id) { if(l <= tree[id].l && r >= tree[id].r)return tree[id].a; int mid = (tree[id].l + tree[id].r)/2; int ans1 = 0,ans2 = 0; if(l <= mid) ans1 = query(l,r,id<<1); if(r > mid) ans2 = query(l,r,id<<1|1); return ans1 + ans2;}int main() { int n; while(scanf("%d",&n) == 1) { init(); build(1,n,1); int ans = 0; for(int i=0;i