链接：
?id=22649
题意就是求有多少联通的块。（一开始自己脑补成双联通然后死也过不了样例。
这道题目的读入也是挺恶心的。
#pragma comment(linker, "/STACK:10240000,10240000")#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <stack>#include <set>#include <map>#define mod 4294967296#define MAX 0x3f3f3f3f#define lson o<<1, l, m#define rson o<<1|1, m+1, r#define SZ(x) ((int)ans.size())#define MAKE make_pair#define INFL 0x3f3f3f3f3f3f3f3fLL#define mem(a) memset(a, 0, sizeof(a))const double pi = acos(-1.0);const double eps = 1e-9;const int N = 105;const int M = 20005;typedef long long ll;using namespace std;int n, m, T;vector <int> G[N];int pre[N], iscut[N], bcc[N], dfs_clock, bcc_cnt;int dfs(int u, int fa) {int lowu = pre[u] = ++ dfs_clock;int child = 0;for(int i = 0; i < G[u].size(); i++) {int v = G[u][i];if(!pre[v]) {child++;int lowv = dfs(v, u);lowu = min(lowu, lowv);if(lowv >= pre[u]) {iscut[u] = 1;}} else if(pre[v] < pre[u] && v != fa) {lowu = min(lowu, pre[v]);}}if(fa < 0 && child == 1) iscut[u] = 0;return lowu;}void find_scc() {mem(pre);mem(iscut);mem(bcc);dfs_clock = bcc_cnt = 0;for(int i = 0; i < n; i++) {if(pre[i] == 0) dfs(i, -1);}}int f[N];int Find(int x) {return f[x] == x ? x : f[x] = Find(f[x]);}int main(){//freopen("in.txt","r",stdin);cin >> T; getchar();getchar();int ca = 1;while(T–) {string s;getline(cin, s);n = s[0]-'A'+1;for(int i = 0; i < n; i++) G[i].clear();for(int i = 0; i < n; i++) {f[i] = i;}while(getline(cin, s) && s != "") {int x = s[0]-'A';int y = s[1]-'A';int fa = Find(x);int fb = Find(y);if(fa != fb) {f[fa] = fb;}}int cnt = 0;for(int i = 0; i < n; i++) {if(f[i] == i) cnt++;}if(ca > 1) puts(""); ca++;cout << cnt << endl;}return 0;}
，有本钱耍个性，离开睁眼闭眼看见的城市，逃离身边的纷纷扰扰，
