此题让我知道了快读和取消同步后的cin貌似不能一起用？

算了，反正取消同步后的cin也挺快的。

 

代码实现如下：

#include 
      
       using namespace std;#define rep(i, a, b) for (register int i = (a); i <= (b); i++)const int maxm = 1e2 + 5;;int n, v, ans = 1;int dp[maxm][maxm];bool vis[maxm][maxm];string s, s1, s2;map
       
         map_city;int MAX(int a, int b) {
    return a > b ? a : b;}void write(int x) {    if (x < 0) {        putchar('-');        x = -x;    }    if (x > 9) write(x / 10);    putchar(x % 10 + '0');}int main() {    std::ios::sync_with_stdio(false);    std::cin.tie(0);    cin >> n >> v;    rep(i, 1, n) {        cin >> s;        map_city[s] = i;    }    rep(i, 1, v) {        cin >> s1 >> s2;        vis[map_city[s1]][map_city[s2]] = 1;        vis[map_city[s2]][map_city[s1]] = 1;    }    dp[1][1] = 1;    rep(i, 1, n - 1)        rep(j, i + 1, n)//i,j之后会对换,所以j就直接从i + 1开始.             rep(k, 1, j - 1)                if (vis[j][k] && dp[i][k]) dp[i][j] = dp[j][i] = MAX(dp[i][j], dp[i][k] + 1);        rep(i, 1, n)        if (vis[i][n]) ans = MAX(ans, dp[i][n]);    write(ans);    return 0;}