LightOJ - 1051 Good or Bad（扫描）

题目大意：给你一个字符串，如果出现三个相邻的辅音字母，或者三个相邻的原音字母，这个字符串就是BAD的，反之就是good。字符串中有？，允许将？变成26个字母中的任意一个。如果字符串能变化成BAD和GOOD，那么就输出MIXED

解题思路：先不考虑？，看是否是BAD，如果是BAD就不用再考虑了
再考虑一下所有？变成原因和辅音的两种情况，看是否会变成BAD
最后考虑一下，是否永远都是BAD，如果不是永远都是BAD，就表示字符串可以变GOOD
判断是否永远是BAD，只要判断一下是否?无论怎么变都是bad就可以

#include <cstdio>
#include <cstring>
const int N = 60;

char str[N];
int len, cas = 1;

void init() {
    scanf("%s", str);
    len = strlen(str);
}

void solve() {
    int vow = 0, con = 0;
    bool bad = 0, good = 0, AllBad = 0;
    for (int i = 0; i < len; i++) {
        if (str[i] == 'A' || str[i] == 'E' || str[i] == 'I' || str[i] == 'O' || str[i] == 'U') {
            vow++;
            con = 0;
        }
        else if (str[i] == '?') {
            vow = 0;
            con = 0;
        } 
        else {
            con++;
            vow = 0;
        }

        if (vow == 3 || con == 5) {
            AllBad = bad = 1;
            break;
        }
    }

    if (AllBad) {
        printf("Case %d: BAD\n", cas++);
        return ;
    }

    if (!bad) {
        vow = 0, con = 0;
        for (int i = 0; i < len; i++) {
            if (str[i] == 'A' || str[i] == 'E' || str[i] == 'I' || str[i] == 'O' || str[i] == 'U' || str[i] == '?') {
                vow++;
                con = 0;
            }
            else {
                vow = 0;
                con++;
            }
            if (vow == 3) {
                bad = true;
                break;
            }
        }
    }

    if (!bad) {
        vow = 0, con = 0;
        for (int i = 0; i < len; i++) {
            if (str[i] == 'A' || str[i] == 'E' || str[i] == 'I' || str[i] == 'O' || str[i] == 'U') {
                vow++;
                con = 0;
            }
            else {
                con++;
                vow = 0;
            }
            if (con == 5) {
                bad = true;
                break;
            }
        }
    }

    for (int i = 0; i < len; i++) {
        if (str[i] == '?') {
            int leftVow = 0, leftCon = 0, rightVow = 0, rightCon = 0;
            for (int j = i - 1; j >= 0; j--) {
                if (str[j] == '?') break;
                if (str[j] == 'A' || str[j] == 'E' || str[j] == 'I' || str[j] == 'O' || str[j] == 'U') {
                    if (leftCon) break;
                    leftVow++;
                }
                else {
                    if (leftVow) break;
                    leftCon++;
                }
            } 

            for (int j = i + 1; j < len; j++) {
                if (str[j] == '?') break;
                if (str[j] == 'A' || str[j] == 'E' || str[j] == 'I' || str[j] == 'O' || str[j] == 'U') {
                    if (rightCon) break;
                    rightVow++;
                }
                else {
                    if (rightVow) break;
                    rightCon++;
                }
            } 

            if ((leftVow == 2 && rightCon == 4 )|| (leftCon == 4 && rightVow == 2)) AllBad = 1;
            if (leftVow == 2 || rightVow == 2) str[i] = 'Q';
            if (leftCon == 4 || rightCon == 4) str[i] = 'A';
            if (AllBad) break;
        }
    }
    if (AllBad) printf("Case %d: BAD\n", cas++);
    else if (bad) printf("Case %d: MIXED\n", cas++);
    else if (!bad) printf("Case %d: GOOD\n", cas++);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}