FFF at Valentine
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 575 Accepted Submission(s): 281
At Valentine's eve, Shylock and Lucar were enjoying their time as any other couples. Suddenly, LSH, Boss of FFF Group caught both of them, and locked them into two separate cells of the jail randomly. But as the saying goes: There is always a way out , the lovers made a bet with LSH: if either of them can reach the cell of the other one, then LSH has to let them go.
The jail is formed of several cells and each cell has some special portals connect to a specific cell. One can be transported to the connected cell by the portal, but be transported back is impossible. There will not be a portal connecting a cell and itself, and since the cost of a portal is pretty expensive, LSH would not tolerate the fact that two portals connect exactly the same two cells.
As an enthusiastic person of the FFF group, YOU are quit curious about whether the lovers can survive or not. So you get a map of the jail and decide to figure it out.
∙Input starts with an integer T (T≤120), denoting the number of test cases.
∙For each case,
First line is two number n and m, the total number of cells and portals in the jail.(2≤n≤1000,m≤6000)
Then next m lines each contains two integer u and v, which indicates a portal from u to v.
If the couple can survive, print “I love you my love and our love save us!”
Otherwise, print “Light my fire!”
题目链接：HDU 6165
似乎比较模版的一道题目，一开始以为是一旦某个缩点后的点的出度超过2就不行了，实际上是可以的，比如1->2、2->3、1->3，这样是三个连通分量且1的出度为2，但是任意取两个点$a,b$还是可以从$a$到达$b$或者$b$到达$a$的，因此更进一步应该是考虑拓扑序上是否同时存在两个可行的点，如果存在说明可以走分岔路这样一来至少分岔路上的两个点就是无法到达的
代码：
#include <stdio.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 10010;
const int M = 60010;
struct edge
{
int to, nxt;
} E[M], e[M];
int head[N], tot;
int dfn[N], low[N], st[N], ins[N], scc, in[N], belong[N], ts, top;
int n, m;
int H[N], Tot;
void init()
{
CLR(head, -1);
tot = 0;
CLR(dfn, 0);
CLR(low, 0);
CLR(ins, 0);
scc = 0;
CLR(in, 0);
ts = top = 0;
CLR(H, -1);
Tot = 0;
}
inline void add(int s, int t)
{
E[tot].to = t;
E[tot].nxt = head[s];
head[s] = tot++;
}
inline void Add(int s, int t)
{
e[Tot].to = t;
e[Tot].nxt = H[s];
H[s] = Tot++;
}
void Tarjan(int u)
{
dfn[u] = low[u] = ++ts;
ins[u] = 1;
st[top++] = u;
int v;
for (int i = head[u]; ~i; i = E[i].nxt)
{
v = E[i].to;
if (!dfn[v])
{
Tarjan(v);
low[u] = min(low[u], low[v]);
}
else if (ins[v])
low[u] = min(low[u], dfn[v]);
}
if (low[u] == dfn[u])
{
++scc;
do
{
v = st[--top];
ins[v] = 0;
belong[v] = scc;
} while (u != v);
}
}
int solve()
{
queue<int>Q;
int i, j;
for (i = 1; i <= n; ++i)
if (!dfn[i])
Tarjan(i);
for (i = 1; i <= n; ++i)
{
for (j = head[i]; ~j; j = E[j].nxt)
{
int v = E[j].to;
if (belong[v] == belong[i])
continue;
++in[belong[v]];
Add(belong[i], belong[v]);
}
}
for (i = 1; i <= scc; ++i)
if (!in[i])
Q.push(i);
while (!Q.empty())
{
if (Q.size() >= 2)
return 0;
int u = Q.front();
Q.pop();
for (int i = H[u]; ~i; i = e[i].nxt)
{
int v = e[i].to;
if (--in[v] == 0)
Q.push(v);
}
}
return 1;
}
int main(void)
{
int T, a, b;
scanf("%d", &T);
while (T--)
{
init();
scanf("%d%d", &n, &m);
while (m--)
{
scanf("%d%d", &a, &b);
add(a, b);
}
solve() ? puts("I love you my love and our love save us!") : puts("Light my fire!");
}
return 0;
}