lca求公共祖先，一般使用倍增法

 1 vector<int>son[N];
 2 int fa[N][35];
 3 int n, m;
 4 int dep[N];
 5 void dfs(int x, int pre){
 6     fa[x][0] = pre;
 7     dep[x] = dep[pre] + 1;
 8     for (int i = 1; (1 << i) <= n; ++i){
 9         fa[x][i] = fa[fa[x][i - 1]][i-1];
10     }
11     for (int a : son[x]){
12         if (a!=pre)
13         dfs(a, x);
14     }
15 }
16 int up(int x, int step){
17     int ret = x;
18     for (int i = 0; (1 << i) <= n; ++i){
19         if (((1 << i)&step)!= 0){
20             ret = fa[ret][i];
21         }
22     }
23     return ret;
24 }
25 int lca(int x, int y){
26     if (dep[x]<dep[y]){
27         swap(x, y);
28     }
29     x = up(x, dep[x] - dep[y]);
30     if (x == y)
31         return x;
32     for (int i = 30; i >= 0; --i){
33         if (fa[x][i] != fa[y][i]){
34             x = fa[x][i];
35             y = fa[y][i];
36         }
37     }
38     return fa[x][0];
39 }
40 int main(){
41     int s;
42     cin >> n >>m>>s;
43     for (int i = 0; i < n - 1; ++i){
44         int a, b;
45         cin >> a >> b;
46         son[a].push_back(b);
47         son[b].push_back(a);
48     }
49     dfs(s,0);//无根树会因为取不同的根节点祖先会发生改变
50     while (m--){
51         int a, b;
52         cin >> a >> b;
53         cout << lca(a, b) << endl;
54     }
55     return 0;
56 }//倍增法求lca

kmp一种快速匹配字符串的方式，一般不会用，我们一般会用他的nextarr数组去求解问题

 1 int nextarr[N];
 2 void getnext(string s){
 3     memset(nextarr, -1, sizeof(nextarr));
 4     nextarr[1] = 0;
 5     int i = 2;
 6     int dex = 0;
 7     int len = s.size();
 8     while (i <=len){
 9         if (s[dex] == s[i - 1]){
10             nextarr[i++] = ++dex;
11         }
12         else if (dex>0){
13             dex = nextarr[dex];
14         }
15         else
16             nextarr[i++] = 0;
17     }
18 }
19 int kmp(string s1, string s2){
20     if (s1.empty() || s2.empty() || s1.size() < s2.size())
21         return -1;
22     int len1 = s1.size();
23     int len2 = s2.size();
24     int i1 = 0;
25     int i2 = 0;
26     while (i1 < len1&&i2 < len2){
27         if (s1[i1] == s2[i2]){
28             i1++;
29             i2++;
30         }
31         else if (i2 == 0)
32             i1++;
33         else
34             i2 = nextarr[i2];
35     }
36     return i2 == len2 ? i1 - i2 : -1;
37 }