题目链接：
http://poj.org/problem?id=2762
|
Going from u to v or from v to u?
Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either Input The first line contains a single integer T, the number of test cases. And followed T cases. The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly. Output The output should contain T lines. Write ‘Yes‘ if the cave has the property stated above, or ‘No‘ otherwise. Sample Input 1 3 3 1 2 2 3 3 1 Sample Output Yes Source POJ Monthly--2006.02.26,zgl & twb |
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题目意思：
给一幅图，判断任意两点v,u是否可到达.(u->v或v->u)都可以。
解题思路：
tarjan+dfs
先求有向图强连通分量，然后缩点建图，统计入度为0的联通分量个数，超过1肯定不行。然后对搜索子树dfs，如果某一节点有超过一个儿子，则这两个儿子之间不能到达，不行。
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 1100
int low[Maxn],dfn[Maxn],sc,bc,sta[Maxn];
int n,m,dep,dei[Maxn],in[Maxn];
bool iss[Maxn];
vector<vector<int> >myv;
vector<vector<int> >tree;
bool hae[Maxn][Maxn];
int ans;
void tarjan(int cur)
{
int ne;
low[cur]=dfn[cur]=++dep;
sta[++sc]=cur;
iss[cur]=true;
for(int i=0;i<myv[cur].size();i++)
{
ne=myv[cur][i];
if(!dfn[ne])
{
tarjan(ne);
if(low[ne]<low[cur])
low[cur]=low[ne];
}
else if(iss[ne]&&dfn[ne]<low[cur])
low[cur]=dfn[ne];
}
if(low[cur]==dfn[cur])
{
++bc;
do
{
ne=sta[sc--];
iss[ne]=false;
in[ne]=bc;
}while(ne!=cur);
}
}
void solve()
{
dep=sc=bc=0;
memset(iss,false,sizeof(iss));
memset(dfn,0,sizeof(dfn));
for(int i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
}
void dfs(int cur)
{
if(ans>2)
return ;
int res=0;
for(int i=0;i<tree[cur].size();i++)
{
int ne=tree[cur][i];
res++;
dfs(ne);
}
if(res>=2)
ans=INF;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
myv.clear();
myv.resize(n+1);
memset(dei,0,sizeof(dei));
for(int i=1;i<=m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
myv[a].push_back(b);
}
solve();
tree.clear();
tree.resize(bc+1);
memset(hae,false,sizeof(hae));
for(int i=1;i<=n;i++)
{
for(int j=0;j<myv[i].size();j++)
{
int ne=myv[i][j];
if(in[i]!=in[ne])
{
dei[in[ne]]++;
if(!hae[in[i]][in[ne]])
{
hae[in[i]][in[ne]]=true;
tree[in[i]].push_back(in[ne]);
}
}
}
}
ans=0;
for(int i=1;i<=bc;i++)
if(!dei[i])
{
ans++;
dfs(i);
if(ans>1)
break;
}
if(ans==1)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}