拓扑排序/DFS HDOJ 4324 Triangle LOVE

 

题目传送门

题意：判三角恋(三元环)。如果A喜欢B，那么B一定不喜欢A，任意两人一定有关系连接

分析：正解应该是拓扑排序判环，如果有环，一定是三元环，证明。 DFS：从任意一点开始搜索，搜索过的点标记，否则超时。看是否存在两点路程只差等于2，如果存在，则说明有上述的三角环。其他做法。

收获：DFS搜索一定要用vis数组啊，否则很容易超时的

 

代码(拓扑排序)：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-25 19:24:24
* File Name     :E_topo.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
char s[N][N];
vector<int> G[N];
int in[N];
int n;

bool Topo_Sort(void)    {
    memset (in, 0, sizeof (in));
    for (int i=1; i<=n; ++i)    {
        for (int j=0; j<G[i].size (); ++j)  in[G[i][j]]++;
    }
    queue<int> Q;   int cnt = 0;
    for (int i=1; i<=n; ++i)    if (!in[i]) Q.push (i);
    while (!Q.empty ()) {
        int u = Q.front (); Q.pop ();
        cnt++;
        for (int i=0; i<G[u].size (); ++i)    {
            int v = G[u][i];
            if (!(--in[v])) Q.push (v);
        }
    }
    if (cnt == n)   return false;
    else    return true;
}

int main(void)    {
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d", &n);
		for (int i=1; i<=n; ++i)    G[i].clear ();
        for (int i=1; i<=n; ++i)	{
			scanf ("%s", s[i] + 1);
            for (int j=1; j<=n; ++j)    {
                if (s[i][j] == '1') G[i].push_back (j);
            }
		}

		printf ("Case #%d: %s\n", ++cas, (Topo_Sort () ? "Yes" : "No"));
	}

    return 0;
}

 

代码(DFS)：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-25 9:44:18
* File Name     :E.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
char s[N][N];
bool vis[N];
int d[N];
int n;

bool DFS(int u)	{
    for (int i=1; i<=n; ++i)    {
        if (s[u][i] == '1') {
            if (vis[i] && d[u] == d[i] + 2) return true;
            else if (!vis[i])   {
                vis[i] = true;  d[i] = d[u] + 1;
                if (DFS (i))    return true;
            }
        }
    }
    return false;
}

int main(void)    {
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d", &n);
		memset (vis, false, sizeof (vis));
		memset (d, 0, sizeof (d));
        for (int i=1; i<=n; ++i)	{
			scanf ("%s", s[i] + 1);
		}

        vis[1] = true;
		printf ("Case #%d: %s\n", ++cas, (DFS (1) ? "Yes" : "No"));
	}

    return 0;
}