Codeforces 761C Dasha and Password(枚举+贪心)

题目链接 Dasha and Password

题目保证一定有解。

考虑到最多只有两行的指针需要移动，那么直接预处理出该行移动到字母数字或特殊符号的最小花费。

然后O(N^3)枚举求最小值即可。

时间复杂度O(N*M+N^3)

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 #define REP(i,n)                for(int i(0); i <  (n); ++i)
 6 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 7 #define dec(i,a,b)              for(int i(a); i >= (b); --i)
 8 #define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
 9 #define INF            1    <<    30
10
11 const int N     =    100000      +       10;
12 const int M     =    10000       +       10;
13 const int Q     =    100       +       10;
14 const int A     =    30          +       1;
15
16
17 int a[Q][Q], c[Q][10], v[Q][10]; char s[Q];
18 int n, m, ans;
19
20 int main(){
21
22     scanf("%d%d", &n, &m);
23     rep(i, 1, n){
24         scanf("%s", s + 1);
25         rep(j, 1, m){
26             if (s[j] >= ‘0‘ && s[j] <= ‘9‘) a[i][j] = 1;
27             else if (s[j] >= ‘a‘ && s[j] <= ‘z‘) a[i][j] = 2;
28             else if (s[j] == ‘&‘ || s[j] == ‘*‘ || s[j] == ‘#‘) a[i][j] = 3;
29         }
30     }
31
32     rep(i, 1, n) rep(j, 1, 3) c[i][j] = INF;
33
34     rep(i, 1, n){
35         rep(j, 1, m){
36             v[i][a[i][j]] = 1;
37             c[i][a[i][j]] = min(c[i][a[i][j]], min(abs(j - 1), abs(m - j) + 1));
38         }
39     }
40
41
42     ans = INF;
43     rep(i, 1, n - 2){
44         rep(j, i + 1, n - 1){
45             rep(k, j + 1, n){
46                 if (v[i][1] && v[j][2] && v[k][3]) ans = min(ans, c[i][1] + c[j][2] + c[k][3]);
47                 if (v[i][1] && v[j][3] && v[k][2]) ans = min(ans, c[i][1] + c[j][3] + c[k][2]);
48                 if (v[i][2] && v[j][1] && v[k][3]) ans = min(ans, c[i][2] + c[j][1] + c[k][3]);
49                 if (v[i][2] && v[j][3] && v[k][1]) ans = min(ans, c[i][2] + c[j][3] + c[k][1]);
50                 if (v[i][3] && v[j][1] && v[k][2]) ans = min(ans, c[i][3] + c[j][1] + c[k][2]);
51                 if (v[i][3] && v[j][2] && v[k][1]) ans = min(ans, c[i][3] + c[j][2] + c[k][1]);
52             }
53         }
54     }
55
56     printf("%d\n", ans);
57     return 0;
58
59 }