CodeChef ISCC2017 HISTOSIM

题目链接

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分析

其实就是字符映射，如果本来就相等，自然Yes。否则，若对应映射关系满足且至少有一个字符未被映射，则为Yes，否则为No

代码

#define maxn 1010char a[maxn], b[maxn];int v[30], d[30];int main(){/*    #ifdef ONLINE_JUDGE        freopen("aplusb.in", "r", stdin);        freopen("aplusb.out", "w", stdout);    #endif*/    // ios::sync_with_stdio(false);    int t;    scanf("%d", &t);    while (t--) {        scanf("%s%s", a, b);        if (!strcmp(a, b)) {            puts("YES");            continue;        }        int x = strlen(a), fg = 1;        for (int i = 1; i < x; i++) {            if (b[i] == b[i - 1]) {                if (a[i] != a[i - 1]) {                    fg = 0;                    break;                }             }        }        if (fg) {            for (int i = 0; i < x; i++) {                v[b[i] - 'a'] = a[i] - 'a';            }            for (int i = 0; i < x; i++) {                if (a[i] != v[b[i] - 'a'] + 'a') {                    fg = 0;                    break;                }            }            if (fg) {                memset(d, 0, sizeof d);                for (int i = 0; i < x; i++) {                    d[b[i] - 'a'] = 1;                }                fg = 0;                for (int i = 0; i < 26; i++) {                    if (!d[i]) {                        fg = 1;                        break;                    }                }                if (fg) {                    puts("YES");                } else {                    puts("NO");                }            } else {                puts("NO");            }        } else {            puts("NO");        }    }}