题目传送门
1 /*
2 题意：查询x的id，每次前排的树倒下
3 使用lower_bound ()查找高度，f[i]记录第一棵高度为x树的位置，查询后+1(因为有序)
4 */
5 #include <cstdio>
6 #include <algorithm>
7 #include <cstring>
8 using namespace std;
9
10 const int MAXN = 1e5 + 10;
11 const int INF = 0x3f3f3f3f;
12 struct A
13 {
14 int v, id;
15 bool operator < (const A &x) const
16 {
17 if (v == x.v) return id < x.id;
18 return v < x.v;
19 }
20 }a[MAXN];
21 int h[MAXN], f[MAXN];
22
23 inline int read(void)
24 {
25 int x = 0, f = 1; char ch = getchar ();
26 while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar ();}
27 while (ch >='0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar ();}
28 return f * x;
29 }
30
31 int main(void) //HDOJ 5233 Gunner II
32 {
33 int n, q;
34 while (scanf ("%d%d", &n, &q) == 2)
35 {
36 for (int i=1; i<=n; ++i) {a[i].v = read (); a[i].id = i;}
37 sort (a+1, a+1+n);
38 for (int i=1; i<=n; ++i)
39 {
40 f[i] = i; h[i] = a[i].v;
41 }
42 while (q--)
43 {
44 int x, p; scanf ("%d", &x);
45 p = lower_bound (h+1, h+1+n, x) - h;
46 if (h[f[p]] != x) {puts ("-1"); continue;}
47 printf ("%d
", a[f[p]].id);
48 f[p]++;
49 }
50 }
51
52 return 0;
53 }
54
55 /*
56 5 5
57 1 2 3 4 1
58 1 3 1 4 2
59 */