2019杭电多校第四场hdu6621 K-th Closest Distance(二分答案+主席树)
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K-th Closest Distance
解题思路
二分答案+主席树
先建主席树,然后二分答案mid,在l和r的区间内查询[p-mid, p+mid]的范围内的数的个数,如果大于k则说明这个范围内存在第k小的数,r=mid,否则不存在,l=mid+1。
代码如下
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int read()
int res = 0, w = 0; char ch = 0;
while(!isdigit(ch))
w |= ch == '-', ch = getchar();
while(isdigit(ch))
res = (res << 3) + (res << 1) + (ch ^ 48);
ch = getchar();
return w ? -res : res;
const int N = 1000005;
int rt[N], cnt;
struct T
int l, r;
int lch, rch;
int sum;
tree[N * 30];
inline int build(int l, int r)
int p = ++cnt;
tree[p].l = l;
tree[p].r = r;
tree[p].lch = tree[p].rch = tree[p].sum = 0;
if(l == r)
return p;
int mid = (l + r) / 2;
tree[p].lch = build(l, mid);
tree[p].rch = build(mid + 1, r);
return p;
inline int insert(int k, int x)
int p = ++cnt;
tree[p].l = tree[k].l;
tree[p].r = tree[k].r;
tree[p].lch = tree[k].lch;
tree[p].rch = tree[k].rch;
tree[p].sum = tree[k].sum + 1;
if(tree[k].l == tree[k].r)
return p;
int mid = (tree[k].l + tree[k].r) / 2;
if(x <= mid)
tree[p].lch = insert(tree[k].lch, x);
else
tree[p].rch = insert(tree[k].rch, x);
return p;
int query(int k1, int k2, int l, int r)
if(tree[k1].l >= l && tree[k1].r <= r)
return tree[k2].sum - tree[k1].sum;
int mid = (tree[k1].l + tree[k1].r) / 2;
int m1 = 0, m2 = 0;
if(l <= mid)
m1 = query(tree[k1].lch, tree[k2].lch, l, r);
if(r > mid)
m2 = query(tree[k1].rch, tree[k2].rch, l, r);
return m1 + m2;
int main()
int _;
scanf("%d", &_);
while(_ --)
int n, m;
scanf("%d%d", &n, &m);
rt[0] = build(1, 1e6);
for(int i = 1; i <= n; i ++)
int x = read();
rt[i] = insert(rt[i -1], x);
int x = 0;
for(int i = 1; i <= m; i ++)
int l, r, p, k;
l = read(), r = read(), p = read(), k = read();
l ^= x, r ^= x, p ^= x, k ^= x;
int ll = 0, rr = 1e6;
int mid = (ll + rr) / 2;
while(ll < rr)
if(query(rt[l - 1], rt[r], max(1, p - mid), min(p + mid, 1000000)) >= k)
rr = mid;
else
ll = mid + 1;
mid = (ll + rr) / 2;
x = mid;
printf("%d\n", x);
return 0;
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