【BZOJ4956】Secret Chamber at Mount Rushmore

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=4956

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暂时结束DP的学习，开始图论的复习了，哦，复习，呵呵呵。。。

这是比较水的一道题，我们只关注从一个点是否有路径可以到达另一个点，而且数据范围又不大，显然是Floyd算法的应用，求传递闭包。

但是，好好读题，把题目的要求在代码当中体现得淋漓尽致，别问我为什么要说这个。。。

#include <cstdio>
#include <cstring>

const int maxa = 30, maxn = 55;

int f[maxa][maxa];
char w1[maxn], w2[maxn];

int main() {
    int m, n;
    char a, b;
    scanf("%d%d", &m, &n);
    for (int i = 1; i <= m; ++i) {
        a = getchar();
        while (a < 'a' || a > 'z') a = getchar();
        b = getchar();
        while (b < 'a' || b > 'z') b = getchar();
        a = a - 'a' + 1, b = b - 'a' + 1;
        f[(int)a][(int)b] = 1;
    }
    for (int i = 1; i <= 26; ++i) f[i][i] = 1;
    for (int k = 1; k <= 26; ++k)
        for (int i = 1; i <= 26; ++i)
            for (int j = 1; j <= 26; ++j)
                f[i][j] = f[i][j] || (f[i][k] && f[k][j]);
    for (int i = 1; i <= n; ++i) {
        scanf("%s%s", w1, w2);
        int w1l = strlen(w1), w2l = strlen(w2), flag = 1;
        if (w1l != w2l) {
            printf("no\n");
            continue;
        }
        for (int j = 0; j < w1l; ++j)
            if (!f[(int)w1[j] - 'a' + 1][(int)w2[j] - 'a' + 1]) {
                printf("no\n");
                flag = 0;
                break;
            }
        if (flag) printf("yes\n");
    }
    return 0;
}