题意 就是求区间第k大，区间 不互相包含。
尝试用treap解决一下 第k大的问题。
1 #include <set> 2 #include <map> 3 #include <cmath> 4 #include <ctime> 5 #include <queue> 6 #include <stack> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstdlib> 11 #include <cstring> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 typedef unsigned long long ull; 16 typedef long long ll; 17 const int inf = 0x3f3f3f3f; 18 const double eps = 1e-8; 19 const int maxn = 1e5+10; 20 struct Node 21 { 22 Node *ch[2]; 23 int key, siz, r; 24 Node (int val) 25 { 26 ch[0] = ch[1] = NULL; 27 siz = 1; 28 key = val; 29 r = rand(); 30 } 31 bool operator < (const Node &rhs)const 32 { 33 return r < rhs.r; 34 } 35 int cmp(int x)const 36 { 37 if (key == x) 38 return -1; 39 return key > x ? 0 : 1; 40 } 41 void maintain() 42 { 43 siz = 1; 44 if (ch[0] != NULL) 45 siz += ch[0] -> siz; 46 if (ch[1] != NULL) 47 siz += ch[1] -> siz; 48 } 49 }; 50 void rotate(Node* &root, int d) 51 { 52 Node *tmp = root -> ch[d^1]; 53 root -> ch[d^1] = tmp -> ch[d]; 54 tmp -> ch[d] = root; 55 root -> maintain(); 56 tmp -> maintain(); 57 root = tmp; 58 } 59 void insert (Node* &root, int x) 60 { 61 if (root == NULL) 62 root = new Node (x); 63 else 64 { 65 int d = x < root -> key ? 0 : 1; 66 insert(root -> ch[d], x); 67 if (root -> ch[d] -> r > root -> r) 68 rotate(root, d^1); 69 } 70 root -> maintain(); 71 } 72 void dele(Node* &root, int x) 73 { 74 int d = root -> cmp(x); 75 if (d == -1) 76 { 77 Node* tmp = root; 78 if (root -> ch[0] != NULL && root -> ch[1] != NULL) 79 { 80 int d1 = (root -> ch[0] -> r > root -> ch[1] -> r ? 1 : 0); 81 rotate(root, d1); 82 dele(root -> ch[d1], x); 83 } 84 else 85 { 86 if (root -> ch[0] == NULL) 87 root = root -> ch[1]; 88 else 89 root = root -> ch[0]; 90 delete tmp; 91 } 92 } 93 else 94 dele(root -> ch[d], x); 95 if (root != NULL) 96 root -> maintain(); 97 } 98 int Get_kth(Node* root,int k) 99 { 100 if (root == NULL || k <= 0 || k > root -> siz) 101 return 0; 102 int s = (root -> ch[0] == NULL ? 0 : root -> ch[0] ->siz); 103 if (s + 1 == k) 104 return root -> key; 105 else if (s >= k) 106 return Get_kth(root -> ch[0], k); 107 else 108 return Get_kth(root -> ch[1], k - s -1); 109 } 110 struct Query 111 { 112 int l, r, k,idx; 113 bool operator < (const Query &rhs)const 114 { 115 return r < rhs.r; 116 //return l < rhs.l || (l == rhs.l&&r < rhs.r); 117 } 118 }Q[50005]; 119 int a[100005], ans[50005]; 120 Node* treap; 121 int main() 122 { 123 #ifndef ONLINE_JUDGE 124 freopen("in.txt","r",stdin); 125 #endif 126 int n, m; 127 while (~scanf ("%d%d",&n, &m)) 128 { 129 for (int i = 0; i < n; i++) 130 scanf ("%d", a+i+1); 131 for (int i = 0; i < m; i++) 132 { 133 scanf ("%d%d%d",&Q[i].l, &Q[i].r, &Q[i].k); 134 Q[i].idx = i; 135 if (Q[i].l > Q[i].r) 136 swap(Q[i].l,Q[i].r); 137 } 138 sort (Q, Q+m); 139 int index = Q[0].l, lidx = 0; 140 for (int i = 0; i < m; i++) 141 { 142 for (int j = lidx; j && j < min(Q[i].l, Q[i-1].r+1); j++) 143 dele(treap, a[j]); 144 lidx = Q[i].l; 145 for (int j = max(Q[i].l,index); j <= Q[i].r; j++) 146 { 147 insert(treap, a[j]); 148 } 149 index = Q[i].r+1; 150 ans[Q[i].idx] = Get_kth(treap, Q[i].k); 151 152 } 153 for (int i = 0; i < m; i++) 154 { 155 printf("%d ",ans[i]); 156 } 157 } 158 return 0; 159 }