hiho_1141

题目

    按顺序给出N个数字，求出所有的逆序对个数（逆序对指数字 Ai > Aj且 i < j) 
题目链接:hiho_1141 
    数据规模为 100000，必须使用O(nlogn)的算法来进行求解。下标i从0到N-1，依次求出数字Ai，在A[0, i-1]中比Ai大的数字个数K，将所有的K进行加和即可得到结果。 
    这种动态的排序+统计，使用treap。

实现

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#include<stdio.h> #include<string.h> #include<iostream> #include<string> #include<set> #include<map> #include<vector> #include<queue> #include<stack> #include<unordered_map> #include<unordered_set> #include<algorithm> using namespace std; struct Node{     int val;     int priority;     int count;     int sum;     Node* childs[2];     Node(int v = 0) :val(v){         priority = rand();         childs[0] = childs[1] = NULL;         count = sum = 1;     }     void Update(){         sum = count;         if (childs[0])             sum += childs[0]->sum;         if (childs[1])             sum += childs[1]->sum;     } }; struct Treap{     Node* root;     Treap(){         root = NULL;     }     void Rotate(Node*& node, bool dir){         Node* ch = node->childs[dir];         node->childs[dir] = ch->childs[!dir];         ch->childs[!dir] = node;         node->Update();         node = ch;     }     void Insert(Node*& node, int val){         if (!node){             node = new Node(val);             return;         }         if (node->val == val){             node->count++;             node->sum++;             return;         }         else{             bool dir = node->val < val;             Insert(node->childs[dir], val);             if (node->childs[dir]->priority > node->priority){                 Rotate(node, dir);             }             node->Update();         }     }     int Search(Node* node, int val){         if (!node)             return 0;         if (node->val == val){             if (node->childs[1])                 return node->childs[1]->sum;             else                 return 0;         }         else if (node->val > val){             return Search(node->childs[0], val) + (node->childs[1]? node->childs[1]->sum + node->count : node->count);         }         else             return Search(node->childs[1], val);             } }; int main(){     Treap treap;     int N;     scanf("%d", &N);     long long int count = 0;     int val;     for (int i = 0; i < N; i++){         scanf("%d", &val);         count += (treap.Search(treap.root, val));         treap.Insert(treap.root, val);     }     printf("%lld\n", count);     return 0; }