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)

#include <bits/stdc++.h>

using namespace std;

#define REP(i,n)                for(int i(0); i <  (n); ++i)
#define rep(i,a,b)              for(int i(a); i <= (b); ++i)
#define dec(i,a,b)              for(int i(a); i >= (b); --i)
#define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
#define INF            1    <<    30

const int N     =    100000      +       10;
const int M     =    10000       +       10;
const int Q     =    100       +       10;
const int A     =    30          +       1;


int a[Q][Q], c[Q][10], v[Q][10]; char s[Q];
int n, m, ans;

int main(){

    scanf("%d%d", &n, &m);
    rep(i, 1, n){
        scanf("%s", s + 1);
        rep(j, 1, m){
            if (s[j] >= '0' && s[j] <= '9') a[i][j] = 1;
            else if (s[j] >= 'a' && s[j] <= 'z') a[i][j] = 2;
            else if (s[j] == '&' || s[j] == '*' || s[j] == '#') a[i][j] = 3;
        }
    }

    rep(i, 1, n) rep(j, 1, 3) c[i][j] = INF;

    rep(i, 1, n){
        rep(j, 1, m){
            v[i][a[i][j]] = 1;
            c[i][a[i][j]] = min(c[i][a[i][j]], min(abs(j - 1), abs(m - j) + 1));
        }
    }


    ans = INF;
    rep(i, 1, n - 2){
        rep(j, i + 1, n - 1){
            rep(k, j + 1, n){
                if (v[i][1] && v[j][2] && v[k][3]) ans = min(ans, c[i][1] + c[j][2] + c[k][3]);
                if (v[i][1] && v[j][3] && v[k][2]) ans = min(ans, c[i][1] + c[j][3] + c[k][2]);
                if (v[i][2] && v[j][1] && v[k][3]) ans = min(ans, c[i][2] + c[j][1] + c[k][3]);
                if (v[i][2] && v[j][3] && v[k][1]) ans = min(ans, c[i][2] + c[j][3] + c[k][1]);
                if (v[i][3] && v[j][1] && v[k][2]) ans = min(ans, c[i][3] + c[j][1] + c[k][2]);
                if (v[i][3] && v[j][2] && v[k][1]) ans = min(ans, c[i][3] + c[j][2] + c[k][1]);
            }
        }
    }

    printf("%d
", ans);
    return 0;

}