題意
如果一個字符串含有連續的3個元音或者連續的5個輔音那麼他就是bad的,否則他就是good的,字符串中含有’?’,’?’有可能是任何字母,如果因爲問號不確定的話就是mix。
思路
因爲數據很小所以剛開始直接考慮模擬,但是問號的處理很麻煩,換成了dp的做法。
dp的方程還是比較顯然的,但是同樣是mixed不太好判斷。
dp[0/1][i][j]表示在i前第j個位時元音和輔音的情況,最後判斷good即可。
代碼
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1000 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;
char s[55];
int dp[2][55][10];
bool vow(char ch)
{
return (ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U');
}
int solve()
{
int len = strlen(s + 1);
memset(dp, 0, sizeof(dp));
dp[0][0][0] = dp[1][0][0] = 1;
bool good = 0, bad = 0;
for (int i = 1; i <= len; i++)
{
if (vow(s[i]) || s[i] == '?')
{
for (int j = 0; j < 2; j++)
{
dp[0][i][j+1] |= dp[0][i-1][j];
dp[1][i][0] |= dp[0][i-1][j];
}
if (dp[0][i-1][2]) bad = 1;
}
if (!vow(s[i]) || s[i] == '?')
{
for (int j = 0; j < 4; j++)
{
dp[1][i][j+1] |= dp[1][i-1][j];
dp[0][i][0] |= dp[1][i-1][j];
}
if (dp[1][i-1][4]) bad = 1;
}
}
for (int i = 0; i < 6; i++)
good |= (dp[0][len][i] | dp[1][len][i]);
if (good && bad) return 3;
if (good && !bad) return 1;
if (!good && bad) return 2;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
scanf("%d", &T);
for (int ncase = 1; ncase <= T; ncase++)
{
scanf("%s", s + 1);
int flag = solve(); //1good 2bad 3mix
printf("Case %d: ", ncase);
if (flag == 1) puts("GOOD");
else if (flag == 2) puts("BAD");
else if (flag == 3) puts("MIXED");
}
return 0;
}