475B - Strongly Connected City

这个题目很简单，意思就是判断是不是强连通图，可以多次暴力dfs，数据也就400*400的暴力。

当然也可以使用经典的求强连通算法

#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <cassert>
#include <stack>
using namespace std;

vector<int>G[410];
vector<int>rG[410];
vector<int>vs;
bool used[410];
int cmp[410];

void add_edge(int from, int to){
	//cout << from << " " << to << endl;
	G[from].push_back(to);
	rG[to].push_back(from);
}

void dfs(int v){
	used[v] = true;
	for (int i = 0; i < G[v].size();i++)
		if (!used[G[v][i]]) dfs(G[v][i]);
	vs.push_back(v);
}

void rdfs(int v,int k){
	used[v] = true;
	cmp[v] = k;
	for (int i = 0; i < rG[v].size(); i++){
		if (!used[rG[v][i]]) rdfs(rG[v][i], k);
	}
}

bool dir[410][2];
int main(){
	int n, m;
	while (cin >> n>>m){
		memset(used, 0, sizeof(used));
		for (int i = 0; i < 410; i++) G[i].clear(), rG[i].clear();
		vs.clear();
		char c;
		for (int i = 0; i < n; i++){
			cin >> c;
			if (c == '>') dir[i][0] = 1;
			else dir[i][0] = 0;
		}

		for (int i = 0; i < m; i++){
			cin >> c;
			if (c == 'v') dir[i][1] = 1;
			else dir[i][1] = 0;
		}

		for (int i = 0; i < n;i++)
		for (int j = 0; j < m; j++){
			if (dir[i][0] && j + 1 < m) add_edge(i*m + j, i*m + j + 1);
			if (!dir[i][0]&& j) add_edge(i*m + j, i*m + j-1);
			if (dir[j][1] && i + 1 < n) add_edge(i*m + j, i*m + j + m);
			if (!dir[j][1] && i) add_edge(i*m + j, i*m + j -m );
		}
		for (int v = 0; v < m*n-1;v++)
		if (!used[v]) dfs(v);

		memset(used, 0, sizeof(used));
		int k = 0;

		for (int i = vs.size() - 1; i >= 0; i--){
			if (!used[vs[i]]) rdfs(vs[i], k++);
		}
		//cout << k << endl;
		if (k == 1) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
}