hdu 6165 FFF at Valentine（强连通分量缩点+拓扑排序）

题目思路倒是不难，我做的时候画了画样例，就感觉只要这个有向图能被一条单向路径串起来成一个链，也就是根单向的直线，那情侣就赢了。否则，就完了。因为成了一个单向的链后，任意两点，都可以从某一点到达另一点。不过，我不会写代码怎么验证有向图是个单向的链，然后就处于懵逼中。这刚学了tarjan缩点，才来做。

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1010;
const int MAXM = 6010;
struct Edge
{
int to,next;
} edge[MAXM];
int head[MAXN],tot;
//Belong[i]表示点i属于第Belong[i]个强连通分量
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,top;
int scc;
bool Instack[MAXN];

void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}

void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}

void Tarjan(int u)
{
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for(int i = head[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if(!DFN[v])
{
Tarjan(v);
if(Low[u] > Low[v])
Low[u] = Low[v];
}
else if(Instack[v] && Low[u] > DFN[v])
Low[u] = DFN[v];
}
if(Low[u] == DFN[u])
{
scc++;
do
{
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc;
}
while(v != u);
}
}

void solve(int N)
{
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
Index = scc = top = 0;
for(int i = 1; i <= N; ++i)
if(!DFN[i])
Tarjan(i);
}

int degree[MAXN];
vector<int> G[MAXN];
void work(int n, int m)
{
memset(degree,0,sizeof(degree));
for(int i = 1; i <= n; ++i) G[i].clear();
int u,v;
for(int i = 1; i <= n; ++i)
{
for(int j = head[i]; j != -1; j = edge[j].next)
{
u = Belong[i];
v = Belong[edge[j].to];
if(u != v)
{
G[u].push_back(v);
degree[v]++;
}
}
}
int cnt = 0;
int p;
for(int i = 1; i <= scc; ++i)
{
if(degree[i] == 0) ++cnt,p=i;
if(cnt > 1) break;
}
if(cnt > 1)
{
printf("Light my fire!\n");
return;
}
cnt = 0;
queue<int> que;
que.push(p);
bool flag = false;
while(!que.empty())
{
int q = que.front();
que.pop();
cnt = 0;
for(int i = 0; i < G[q].size(); ++i)
{
degree[G[q][i]]--;
if(degree[G[q][i]] == 0)
{
++cnt;
que.push(G[q][i]);
}
}
if(cnt > 1)
{
flag = true;
break;
}
}
if(flag)
printf("Light my fire!\n");
else
printf("I love you my love and our love save us!\n");
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
init();
int n,m,a,b;
scanf("%d %d",&n,&m);
for(int i = 0; i < m; ++i)
{
scanf("%d %d",&a,&b);
addedge(a,b);
}
solve(n);
work(n,m);
}
return 0;
}