二分查找 BestCoder Round #42 1002 Gunner II

 

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\n", 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 */