POJ 3895 Cycles of Lanes

本文主要是介绍POJ 3895 Cycles of Lanes，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题意是求最大环的边数。做这题发现用vector特别好用。DFS即可。

/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 100000000
#define LOCA
#define PI acos(-1.0)
using namespace std;
bool used[4455];
int num[4455];
int maxlen;
vector<int>G[4455];
void dfs(int v, int t)
{int i;num[v] = t;used[v] = true;int x = G[v].size();for(i = 0; i < x; i++){if(!used[G[v][i]]){dfs(G[v][i], t + 1);}else{if(maxlen < num[v] - num[G[v][i]] + 1)maxlen = num[v] - num[G[v][i]] + 1;}}
}
int main()
{
#ifdef LOCALfreopen("ride.in","r",stdin);freopen("ride.out","w",stdout);
#endifint T, n, m, i, j, x, y;scanf("%d", &T);while(T--){memset(used, false, sizeof(used));memset(num, 0, sizeof(num));scanf("%d%d", &n, &m);for(i = 1; i <= n; i++)G[i].clear();for(i = 0; i < m; i++){scanf("%d%d", &x, &y);G[x].push_back(y);G[y].push_back(x);}maxlen = 0;for(i = 1; i <= n; i++){if(!used[i])dfs(i, 1);}if(maxlen > 2) printf("%d\n", maxlen);else puts("0");}return 0;
}

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