SDUT 2604 Thrall’s Dream （单连通的判断）

大意：给定一些有向图间的关系问任意两点是不是可达的。

思路：Tarjan后直接看入度和出度为0的数量关系，如果大于1则肯定是不可能，相连通的。

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1

using namespace std;

const int Ma = 10100;
struct node{
int to,w,next;
}q[Ma];
int head[Ma],dfn[Ma],num[Ma],du[Ma],stk[Ma],vis[Ma],low[Ma];
int cnt,top,tim,scc,out[Ma],in[Ma];

bool mp[1100][1100];
void Add(int a,int b){
q[cnt].to = b;
q[cnt].next = head[a];
head[a] = cnt++;
}
void init(){
scc = cnt = top = 0;
tim = 1;
memset(head,-1,sizeof(head));
memset(dfn,0,sizeof(dfn));
memset(num,0,sizeof(num));
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
memset(vis,0,sizeof(vis));
memset(low,0,sizeof(low));
memset(mp,false,sizeof(mp));
}
void Tarjan(int u){
low[u] = dfn[u] = tim++;
vis[u] = 1;
stk[top++] = u;
for(int i = head[u]; ~i ; i = q[i].next){
int v = q[i].to;
if(!vis[v]){
Tarjan(v);
low[u] = min(low[u],low[v]);
}
else
low[u] = min(low[u],dfn[v]);
}

if(low[u] == dfn[u]){//找到极大联通分量
scc++;
while(top > 0&&stk[top] != u){
top --;
vis[stk[top] ] = 2;
num[stk[top] ] = scc;
}
}

}
int main(){
int n,m,i,j,k,a,b,c,cla;
scanf("%d",&cla);
for(int zu = 1;zu <=cla;++ zu){
scanf("%d%d",&n,&m);
init();
for(i = 0;i <m;++ i){
scanf("%d%d",&a,&b);
Add(a,b);
}
for(i = 1;i <= n;++ i)
if(!dfn[i])
Tarjan(i);
bool bj = false;
for(i = 1;i <= n;++ i){
for(j = head[i];~j; j = q[j].next){
int v = q[j].to;
if(num[i] != num[v]){
out[num[i] ]++;
in[num[v] ]++;
}
}

}
int ans1,ans2;
ans1 = ans2 =0;
for(i = 1;i <= scc;++ i){
if(in[i]==0)
ans1++;
if(out[i]==0)
ans2++;
cout<<i<<endl;
}
printf("Case %d: ",zu);
if(ans1>1||ans2>1){
puts("The Burning Shadow consume us all");
}
else{
puts("Kalimdor is just ahead");
}
}
return 0;
}