思路：我们需要求距离为d+1的LIS，我们可以用延迟更新方法，a[]为该点数值，f[]为以改点结尾的LIS长度。当inx - d - 1 > 0时，我们才更新这个点的信息，这样就解决了距离的问题。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <queue> 5 6 using namespace std; 7 8 #define ll long long 9 #define pb push_back 10 #define lson(rt) (rt << 1) 11 #define rson(rt) (rt << 1 | 1) 12 #define fi first 13 #define se second 14 15 const int N = 1e5 + 10; 16 struct node 17 { 18 int rt, l, r, sum; 19 int mid() { return(l + r) >> 1; } 20 inline void init(int l1, int r1) { l = l1; r = r1; sum = 0; } 21 }tree[N << 2]; 22 int a[N], f[N]; 23 24 void Build(int rt, int l, int r) 25 { 26 //printf("rt = %d l = %d r = %d ", rt, l, r); 27 tree[rt].init(l, r); 28 if(l == r) return ; 29 Build(lson(rt), l, tree[rt].mid()); 30 Build(rson(rt), tree[rt].mid() + 1, r); 31 } 32 33 inline void Push_up(int rt) 34 { 35 tree[rt].sum = max(tree[lson(rt)].sum, tree[rson(rt)].sum); 36 } 37 38 void Update(int rt, int l, int r, int v, int sum) 39 { 40 //cout << "54 lines" << endl; 41 //printf("rt = %d l = %d r = %d v = %d ", rt, l , r, v); 42 if(l == r) { 43 tree[rt].sum = sum; 44 return ; 45 } 46 47 if(tree[rt].mid() >= v) { 48 Update(lson(rt), l, tree[rt].mid(), v, sum); 49 } else { 50 Update(rson(rt), tree[rt].mid() + 1, r, v, sum); 51 } 52 53 Push_up(rt); 54 } 55 56 int Search(int rt, int l, int r, int vl, int vr) 57 { 58 //cout << "34 lines" << endl; 59 if(vl <= l && r <= vr) return tree[rt].sum; 60 61 if(tree[rt].mid() >= vr) { 62 return Search(lson(rt), l, tree[rt].mid(), vl, vr); 63 } else if(tree[rt].mid() < vl) { 64 return Search(rson(rt), tree[rt].mid() + 1, r, vl, vr); 65 } else { 66 return max(Search(lson(rt), l, tree[rt].mid(), vl, vr), 67 Search(rson(rt), tree[rt].mid() + 1, r, vl, vr)); 68 } 69 } 70 71 void solve() 72 { 73 int n, d; 74 while(~scanf("%d%d", &n, &d)) { 75 76 int Max_v = 0; 77 for(int i = 1; i <= n; ++i) { 78 scanf("%d", a + i); 79 Max_v = max(Max_v, a[i]); 80 } 81 Build(1, 0, Max_v); 82 //cout << "75 lines" << endl; 83 //cout << "Max_v = " << Max_v << endl; 84 for(int i = 0; i <= n; ++i) f[i] = 0; 85 //for(int i = 0; i <= Max_v; ++i) printf("%d ", f[i]); 86 //printf(" "); 87 88 int Max_len = 0; 89 for(int i = 1; i <= n; ++i) { 90 //cout << "i = " << i << endl; 91 if(i - d > 1) Update(1, 0, Max_v, a[i - d - 1], f[i - d - 1]); 92 if(a[i] > 0) { 93 f[i] = max(Search(1, 0, Max_v, 0, a[i] - 1) + 1, f[i]); 94 } else f[i] = 1; 95 96 Max_len = max(Max_len, f[i]); 97 } 98 99 //printf("Max_len = %d ", Max_len); 100 printf("%d ", Max_len); 101 } 102 } 103 104 int main() 105 { 106 solve(); 107 108 return 0; 109 } 110 111 112 /* 113 16 3 114 1 1 1 1 2 2 2 2 3 3 3 6 4 6 6 7 115 116 117 17 3 118 1 1 1 1 2 2 2 2 3 3 3 6 6 6 4 4 7 119 120 121 17 3 122 1 1 1 1 2 2 2 2 3 3 3 5 4 4 4 4 7 123 124 125 10 2 126 1 1 1 2 2 6 4 6 6 7 127 128 129 10 3 130 1 1 1 2 2 6 4 6 6 7 131 132 133 7 1 134 1 1 2 6 4 4 7 135 136 137 4 138 5 139 5 140 3 141 4 142 4 143 144 145 */