SRM 608 D2 L3：BigOEasy，DFS

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题目：http://community.topcoder.com/stat?c=problem_statement&pm=13002

看似复杂，其实只要判断图中是否有两个不同的环有至少一个相同的顶点。从不同的顶点出发进行DFS，判断出发点在几个环中。

代码：

#include <algorithm>#include <iostream>#include <sstream>#include <string>#include <vector>#include <stack>#include <deque>#include <queue>#include <set>#include <map>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <cstring>#include <ctime>using namespace std;#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)/*************** Program Begin **********************/bool visit[55];int ct;int cur;class BigOEasy {public:vector <string> graph;int N;void DFS(int v){visit[v] = true;for (int i = 0; i < N; i++) {if (graph[v][i] == 'Y') {if (visit[i]) {if (i == cur) {++ct;}} else {DFS(i);}}}}    string isBounded(vector <string> graph) {this->graph = graph;this->N = graph.size();string res = "Bounded";memset(visit, 0, sizeof(visit));for (int i = 0; i < N; i++) {memset(visit, 0, sizeof(visit));ct = 0;cur = i;DFS(i);if (ct > 1) {res = "Unbounded";break;}}        return res;    }};/************** Program End ************************/

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