hdu-3183A Magic Lamp (RMQ)

Posted 2020-11-26 hfcdyp

hdu-3183A Magic Lamp

题意：对给定m位数删除其中n位，不改变剩余数字排列，求剩余数字组成的最小数。

思路：选m - n个数 RMQ求最小值 t数组求最小值第一次出现的位置 左右区间随之改变

wa了两次 没特判 m = n 的情况 / t数组存了最小值最后一次出现的位置(25行没用小于等于号用了小于号)

 1 #include<map>
 2 #include<set>
 3 #include<cmath>
 4 #include<cstdio>
 5 #include<vector>
 6 #include<cstring>
 7 #include<cstdlib>
 8 #include<iostream>
 9 #include<algorithm>
10 using namespace std;
11 const int N = 20;
12 int n, q, len;
13 int a[1007],b[1007], f[1007][N], t[1007][N];
14 void RMQ()
15 {
16     for (int i = 1; i <= len; i++)
17     {
18         f[i][0] = a[i];
19         t[i][0] = i;
20     }
21     for (int j = 1; (1 << j) <= len; j++)
22     {
23         for (int i = 1; i + (1<<j) - 1 <= len; i++)
24         {
25             if(f[i][j - 1] <= f[i + (1 << (j - 1))][j - 1])
26             {
27                 f[i][j] = f[i][j - 1];
28                 t[i][j] = t[i][j - 1];
29             }
30             else
31             {
32                 f[i][j] = f[i + (1 << (j - 1))][j - 1];
33                 t[i][j] = t[i + (1 << (j - 1))][j - 1];
34             }
35         }
36     }
37 }
38 int main()
39 {
40     string s;
41     while (cin >> s)
42     {
43         cin >> n;
44         len = s.size();
45         if (n == len) cout << "0" << endl;
46         else
47         {
48             for (int i = 0; i < len; i++) a[i + 1] = s[i] - ‘0‘;
49             RMQ();
50             int k = len - n;
51             int l = 1, r = len - k + 1;
52             for (int i = 1; i <= k; i++)
53             {
54                 int s = 0;
55                 while (1 << (s + 1) < (r - l + 1)) s++;
56                 if (f[l][s] <= f[r - (1 << s) + 1][s])
57                 {
58                     b[i] = f[l][s];
59                     l = t[l][s] + 1;
60                     r++;
61                 }
62                 else
63                 {
64                     b[i] = f[r - (1 << s) + 1][s];
65                     l = t[r - (1 << s) + 1][s] + 1;
66                     r++;
67                 }
68             }
69             int i = 1;
70             for (i = 1; i < k; i++)
71             {
72                 if (b[i] != 0) break;
73             }
74             for (int j = i; j <= k; j++) cout << b[j];
75             cout << endl;
76         }
77         
78     }
79 
80 }