LightOJ 1051 - Good or Bad(DP)

题意：给一字符串，包括字母和问号，问号可能是任何英文字母。若出现连续3个元音或5个辅音，则为BAD，否则为GOOD。若既可能为GOOD也可能为BAD，输出"MIXED"。

思路：dp1[i][j]表示到i为止，连续j个元音可否达到。dp2表示辅音。

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
using namespace std;

typedef long long LL;
#define mem(a, n) memset(a, n, sizeof(a))
#define ALL(v) v.begin(), v.end()
#define si(a) scanf("%d", &a)
#define sii(a, b) scanf("%d%d", &a, &b)
#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pb push_back
#define eps 1e-8
const int inf = 0x3f3f3f3f, N = 50 + 5, MOD = 1e9 + 7;

int T, cas = 0;
int n, m;
int dp1
[4], dp2
[6], a
;
char s
;
char print[]
= { "MIXED", "GOOD", "BAD" };
int judge(char x)  {
return x == 'A' || x=='E' || x=='I' || x=='O' || x == 'U';
}

int main(){
#ifdef LOCAL
freopen("/Users/apple/input.txt", "r", stdin);
//	freopen("/Users/apple/out.txt", "w", stdout);
#endif

si(T);
while(T --) {
scanf("%s", s + 1);
mem(a, 0);
n = strlen(s + 1);
mem(dp1, 0), mem(dp2, 0);
dp1[0][0] = dp2[0][0] = 1;
for(int i = 1; i <= n; i ++) {
if(s[i] == '?') a[i] = 0;
else if(judge(s[i])) a[i] = 1;
else a[i] = 2;

for(int j = 0; j <= 2; j ++) {
if(dp1[i-1][j]) {
if(a[i] == 0 || a[i] == 2) dp2[i][1] = 1;
if(a[i] == 0 || a[i] == 1) dp1[i][j+1] = 1;
}
}
for(int j = 0; j <= 4; j ++) {
if(dp2[i-1][j]) {
if(a[i] == 0 || a[i] == 2) dp2[i][j+1] = 1;
if(a[i] == 0 || a[i] == 1) dp1[i][1] = 1;
}
}
}
int good = 0, bad = 0, ans;
for(int i = 0; i <= 2; i ++) good |= dp1
[i];
for(int i = 0; i <= 4; i ++) good |= dp2
[i];
for(int i = 1; i <= n; i ++) if(dp1[i][3] || dp2[i][5]) bad = 1;
if(good && bad) ans = 0;
else if(good) ans = 1;
else ans = 2;

printf("Case %d: %s\n", ++ cas, print[ans]);
}

return 0;
}