有N条链，每条链上有len[i]个点，现在树上有一个白点（根），我们每次可以选择一条链，链接到已知树上的任意白点上，然后链接的两点变黑，我们想知道树上白点数大于等于K的时候，第K近的白点距离根节点的距离。
所以，有贪心的策略，我们每次都选剩下的最长的链挂到树上去，然后每次查询答案即可，而且我们肯定选择链中心的点，去挂在树上距离根节点最近的白点上面。然后就是一个简单的区间更新和区间查询的过程了，线段树的简单操作了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <string> 5 #include <cstring> 6 #include <algorithm> 7 #include <limits> 8 #include <vector> 9 #include <stack> 10 #include <queue> 11 #include <set> 12 #include <map> 13 #include <bitset> 14 #include <unordered_map> 15 #include <unordered_set> 16 #define lowbit(x) ( x&(-x) ) 17 #define pi 3.141592653589793 18 #define e 2.718281828459045 19 #define INF 0x3f3f3f3f 20 #define HalF (l + r)>>1 21 #define lsn rt<<1 22 #define rsn rt<<1|1 23 #define Lson lsn, l, mid 24 #define Rson rsn, mid+1, r 25 #define QL Lson, ql, qr 26 #define QR Rson, ql, qr 27 #define myself rt, l, r 28 #define pii pair<int, int> 29 #define MP(a, b) make_pair(a, b) 30 using namespace std; 31 typedef unsigned long long ull; 32 typedef unsigned int uit; 33 typedef long long ll; 34 const int maxN = 8e5 + 7; 35 int N, K; 36 ll len[maxN]; 37 ll tree[maxN << 3], lazy[maxN << 3]; 38 void pushdown(int rt, int l, int r) 39 { 40 if(lazy[rt]) 41 { 42 int mid = HalF; 43 tree[lsn] += lazy[rt] * (mid - l + 1); 44 tree[rsn] += lazy[rt] * (r - mid); 45 lazy[lsn] += lazy[rt]; 46 lazy[rsn] += lazy[rt]; 47 lazy[rt] = 0; 48 } 49 } 50 void pushup(int rt) { tree[rt] = tree[lsn] + tree[rsn]; } 51 void update(int rt, int l, int r, int ql, int qr, ll val) 52 { 53 if(ql <= l && qr >= r) 54 { 55 lazy[rt] += val; 56 tree[rt] += val * (r - l + 1); 57 return; 58 } 59 pushdown(myself); 60 int mid = HalF; 61 if(qr <= mid) update(QL, val); 62 else if(ql > mid) update(QR, val); 63 else { update(QL, val); update(QR, val); } 64 pushup(rt); 65 } 66 int _Fir(int rt, int l, int r) 67 { 68 if(l == r) return l; 69 pushdown(myself); 70 int mid = HalF; 71 if(tree[lsn]) return _Fir(Lson); 72 else return _Fir(Rson); 73 } 74 int _Ans(int rt, int l, int r, ll kth) 75 { 76 if(l == r) return l; 77 pushdown(myself); 78 int mid = HalF; 79 if(kth <= tree[lsn]) return _Ans(Lson, kth); 80 else return _Ans(Rson, kth - tree[lsn]); 81 } 82 int main() 83 { 84 scanf("%d%d", &N, &K); 85 for(int i = 1; i <= N; i ++) scanf("%lld", &len[i]); 86 sort(len + 1, len + N + 1, greater<ll>() ); 87 int ans = INF; 88 int _UP = 8e5 + 5; 89 update(1, 0, _UP, 0, 0, 1); 90 for(int i = 1, pos, nex_L, nex_R, tmp; i <= N; i ++) 91 { 92 pos = _Fir(1, 0, _UP); 93 update(1, 0, _UP, pos, pos, -1); 94 nex_L = pos + 2; 95 tmp = (int)len[i] - 1; 96 nex_R = nex_L + tmp / 2 - 1; 97 update(1, 0, _UP, nex_L, nex_R, 2); 98 if(tmp & 1) update(1, 0, _UP, nex_R + 1, nex_R + 1, 1); 99 if(tree[1] >= K) 100 { 101 ans = min(ans, _Ans(1, 0, _UP, K)); 102 } 103 } 104 if(ans < INF) printf("%d ", ans); 105 else printf("-1 "); 106 return 0; 107 }