给定一个数组nums[n],如果i<j
并且nums[i]>2*nums[j]
,则称(i,j)是一个重要翻转对。求nums[n]中重要翻转对的数量。
1<=n<=5e4; nums[i]在int范围内
直接套平衡树模板即可。
template <typename TYPE>
struct Treap {
struct Node {
TYPE data, sum;
int rnd, siz, dup, son[2];
void init(const TYPE & d) {
data = sum = d;
rnd = rand();
siz = dup = 1;
son[0] = son[1] = 0;
}
};
Treap(size_t sz, bool multi):multiple(multi) {
node.resize(sz);
reset();
}
int newnode(const TYPE & d) {
total += 1;
node[total].init(d);
return total;
}
void reset() { root = total = 0; }
void maintain(int x) {
node[x].siz = node[x].dup;
node[x].sum = node[x].data * node[x].dup;
if (node[x].son[0]) {
node[x].siz += node[node[x].son[0]].siz;
node[x].sum += node[node[x].son[0]].sum;
}
if (node[x].son[1]) {
node[x].siz += node[node[x].son[1]].siz;
node[x].sum += node[node[x].son[1]].sum;
}
}
void rotate(int d, int &r) {
int k = node[r].son[d^1];
node[r].son[d^1] = node[k].son[d];
node[k].son[d] = r;
maintain(r);
maintain(k);
r = k;
}
void insert(const TYPE &data, int &r, bool &ans) {
if (r) {
if (!(data < node[r].data) && !(node[r].data < data)) {
ans = false;
if (multiple) {
node[r].dup += 1;
maintain(r);
}
} else {
int d = data < node[r].data ? 0 : 1;
insert(data, node[r].son[d], ans);
if (node[node[r].son[d]].rnd > node[r].rnd) {
rotate(d^1, r);
} else {
maintain(r);
}
}
} else {
r = newnode(data);
}
}
void getkth(int k, int r, TYPE& data) {
int x = node[r].son[0] ? node[node[r].son[0]].siz : 0;
int y = node[r].dup;
if (k <= x) {
getkth(k, node[r].son[0], data);
} else if (k <= x + y) {
data = node[r].data;
} else {
getkth(k-x-y, node[r].son[1], data);
}
}
TYPE getksum(int k, int r) {
if (k <= 0 || r == 0) return 0;
int x = node[r].son[0] ? node[node[r].son[0]].siz : 0;
int y = node[r].dup;
if (k <= x) return getksum(k, node[r].son[0]);
if (k <= x+y) return node[node[r].son[0]].sum + node[r].data * (k-x);
return node[node[r].son[0]].sum + node[r].data * y + getksum(k-x-y,node[r].son[1]);
}
void erase(const TYPE& data, int & r) {
if (r == 0) return;
int d = -1;
if (data < node[r].data) {
d = 0;
} else if (node[r].data < data) {
d = 1;
}
if (d == -1) {
node[r].dup -= 1;
if (node[r].dup > 0) {
maintain(r);
} else {
if (node[r].son[0] == 0) {
r = node[r].son[1];
} else if (node[r].son[1] == 0) {
r = node[r].son[0];
} else {
int dd = node[node[r].son[0]].rnd > node[node[r].son[1]].rnd ? 1 : 0;
rotate(dd, r);
erase(data, node[r].son[dd]);
}
}
} else {
erase(data, node[r].son[d]);
}
if (r) maintain(r);
}
int ltcnt(const TYPE& data, int r) {
if (r == 0) return 0;
int x = node[r].son[0] ? node[node[r].son[0]].siz : 0;
if (data < node[r].data) {
return ltcnt(data, node[r].son[0]);
}
if (!(data < node[r].data) && !(node[r].data < data)) {
return x;
}
return x + node[r].dup + ltcnt(data, node[r].son[1]);
}
int gtcnt(const TYPE& data, int r) {
if (r == 0) return 0;
int x = node[r].son[1] ? node[node[r].son[1]].siz : 0;
if (data > node[r].data) {
return gtcnt(data, node[r].son[1]);
}
if (!(data < node[r].data) && !(node[r].data < data)) {
return x;
}
return x + node[r].dup + gtcnt(data, node[r].son[0]);
}
int count(const TYPE& data, int r) {
if (r == 0) return 0;
if (data < node[r].data) return count(data, node[r].son[0]);
if (node[r].data < data) return count(data, node[r].son[1]);
return node[r].dup;
}
void prev(const TYPE& data, int r, TYPE& result, bool& ret) {
if (r) {
if (node[r].data < data) {
if (ret) {
result = max(result, node[r].data);
} else {
result = node[r].data;
ret = true;
}
prev(data, node[r].son[1], result, ret);
} else {
prev(data, node[r].son[0], result, ret);
}
}
}
void next(const TYPE& data, int r, TYPE& result, bool& ret) {
if (r) {
if (data < node[r].data) {
if (ret) {
result = min(result, node[r].data);
} else {
result = node[r].data;
ret = true;
}
next(data, node[r].son[0], result, ret);
} else {
next(data, node[r].son[1], result, ret);
}
}
}
vector<Node> node;
int root, total;
bool multiple;
bool insert(const TYPE& data) {
bool ret = true;
insert(data, root, ret);
return ret;
}
bool kth(int k, TYPE &data) {
if (!root || k <= 0 || k > node[root].siz)
return false;
getkth(k, root, data);
return true;
}
TYPE ksum(int k) {
assert(root && k>0 && k<=node[root].siz);
return getksum(k, root);
}
int count(const TYPE &data) {
return count(data, root);
}
int size() const {
return root ? node[root].siz : 0;
}
void erase(const TYPE& data) {
return erase(data, root);
}
int ltcnt(const TYPE& data) {
return ltcnt(data, root);
}
int gtcnt(const TYPE& data) {
return gtcnt(data, root);
}
bool prev(const TYPE& data, TYPE& result) {
bool ret = false;
prev(data, root, result, ret);
return ret;
}
bool next(const TYPE& data, TYPE& result) {
bool ret = false;
next(data, root, result, ret);
return ret;
}
};
class Solution {
public:
int reversePairs(vector<int>& nums) {
Treap<long long> tp(50005, true);
int ans = 0;
for (auto i : nums) {
ans += tp.gtcnt(2LL * i);
tp.insert(i);
}
return ans;
}
};
标签:node,return,int,data,son,lc493,TYPE,数目,翻转
From: https://www.cnblogs.com/chenfy27/p/18076788