【来源】网上流传的2017-360秋招笔试题
【问题描述】
【算法思路】
直接求解会超时。使用“线段树”，节点信息为当前区段的元素个数。假设输入元素最大值为M, 建立有M个叶子节点的树，再依次插入元素并向上更新节点信息。
【程序】
1 #include <iostream> 2 #include <vector> 3 4 //#pragma comment(linker, "/STACK:102400000,102400000") 5 using namespace std; 6 struct Segment 7 { 8 int l, r; 9 int value; 10 bool lazyTag; 11 int lazyValue; 12 Segment() :l(0), r(0), value(INT_MAX), lazyTag(false) {}; 13 }; 14 void buildTree(vector<Segment> &tree, int node, int l, int r) 15 { 16 tree[node].l = l; 17 tree[node].r = r; 18 if (l == r) return; 19 buildTree(tree, node << 1, l, (l + r) >> 1); 20 buildTree(tree, (node << 1) + 1, (l + r + 1) >> 1, r); 21 tree[node].value = tree[node << 1].value + tree[(node << 1) + 1].value; 22 } 23 void update(vector<Segment> &tree, int node) 24 { 25 if (node == 0) return; 26 // <<1相当于*2，>>1相当于/2取整 27 tree[node].value = tree[node << 1].value+tree[(node << 1) + 1].value; 28 update(tree, node >> 1); 29 } 30 void pushDown(vector<Segment> &tree, int father) 31 { 32 if (tree[father].lazyTag) 33 { 34 int leftChild = father << 1, rightChild = (father << 1) + 1; 35 tree[leftChild].value = (tree[leftChild].r - tree[leftChild].l + 1)*tree[father].lazyValue; 36 tree[leftChild].lazyTag = true; 37 tree[leftChild].lazyValue = tree[father].lazyValue; 38 tree[rightChild].value = (tree[rightChild].r - tree[rightChild].l + 1)*tree[father].lazyValue; 39 tree[rightChild].lazyTag = true; 40 tree[rightChild].lazyValue = tree[father].lazyValue; 41 tree[father].lazyTag = false; 42 } 43 } 44 void updateInterval(vector<Segment> &tree, int node, int l, int r, int value) 45 { 46 if (tree[node].l == l&&tree[node].r == r) 47 { 48 tree[node].lazyTag = true; 49 tree[node].value = (r - l + 1)*value; 50 tree[node].lazyValue = value; 51 return; 52 } 53 else { 54 //pushDown(tree, node); 55 if (tree[node << 1].r >= r) updateInterval(tree, node << 1, l, r, value); 56 else if (tree[(node << 1) + 1].l <= l) updateInterval(tree, (node << 1) + 1, l, r, value); 57 else 58 { 59 updateInterval(tree, node << 1, l, tree[node << 1].r, value); 60 updateInterval(tree, (node << 1) + 1, tree[(node << 1) + 1].l, r, value); 61 } 62 } 63 tree[node].value = tree[node << 1].value + tree[(node << 1) + 1].value; 64 } 65 66 int query(vector<Segment> &tree, int node, int l, int r) 67 { 68 if (tree[node].l == l&&tree[node].r == r) return tree[node].value; 69 else 70 { 71 //pushDown(tree, node); 72 if (tree[node << 1].r >= r) return query(tree, node << 1, l, r); 73 else if (tree[(node << 1) + 1].l <= l) return query(tree, (node << 1) + 1, l, r); 74 else 75 { 76 return query(tree, node << 1, l, tree[node << 1].r) + query(tree, (node << 1) + 1, tree[(node << 1) + 1].l, r); 77 } 78 } 79 } 80 #define MAXN 200005 81 int A[MAXN]; 82 int main() { 83 #ifdef DEBUG 84 ifstream cin("in.txt"); 85 ofstream cout("out.txt"); 86 //freopen("in.txt", "r",stdin); 87 //freopen("out.txt", "w",stdout); 88 #endif 89 int n=0, a, b, size = 1; 90 int N; 91 cin >> N; 92 for (size_t i = 0; i < N; i++) 93 { 94 cin >> A[i]; 95 n = max(n, A[i]); 96 } 97 while (size < n) size <<= 1; 98 vector<Segment> tree(size * 2); 99 for (size_t i = 0; i < size; i++) 100 { 101 tree[i + size].value=0; 102 } 103 buildTree(tree, 1, 1, size); 104 int ans; 105 for (size_t i = 0; i < N; i++) 106 { 107 tree[A[i]-1 + size].value++; 108 update(tree, (A[i]-1 + size)/2); 109 ans = query(tree,1,A[i],size); 110 if (!i) 111 cout << ans - tree[A[i] - 1 + size].value; //减去相等的个数，输出大于的个数 112 else 113 cout<< ' '<< ans - tree[A[i] - 1 + size].value; 114 } 115 116 return 0; 117 }