Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
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 go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair
of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?
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.
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
题意：jiajia有一个洞穴，洞穴中有n个房间，房间的连接是有方向的,jiajia想知道u与v之间是不是有路径u->v或v->u
思路：单连通问题，首先将图中的强连通的部分进行缩点，构成一颗树，接下来拓扑排序，判断拓扑排序的两点之间是不是有边，如果没有边则图不符合要求。
题意：jiajia有一个洞穴，洞穴中有n个房间，房间的连接是有方向的,jiajia想知道u与v之间是不是有路径u->v或v->u
思路：单连通问题，首先将图中的强连通的部分进行缩点，构成一颗树，接下来拓扑排序，判断拓扑排序的两点之间是不是有边，如果没有边则图不符合要求。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
const int Max = 1100;
const int INF = 0x3f3f3f3f;
typedef struct node
{
int v;
int next;
}Line ;
Line Li[Max*6];
int Head[Max],top;
int Map[Max][Max];
int Du[Max],pre[Max];
int vis1[Max],dfn[Max],low[Max];
bool vis2[Max];
int dep;
int Point[Max*5][2],Num;
int topo[Max],num;
int a[Max],ToNum;
stack < int >S;
void AddEdge(int u,int v)
{
Li[top].v = v; Li[top].next = Head[u];
Head[u] = top++;
}
int Find(int x)
{
return pre[x]==-1?x:pre[x] = Find(pre[x]);
}
void Tarjan(int u) //强连通缩点
{
dfn[u] = low[u] = dep++;
vis1[u] = 1;
S.push(u);
for(int i=Head[u];i!=-1;i=Li[i].next)
{
if(vis1[Li[i].v]==1)
{
low[u]=min(low[u],dfn[Li[i].v]);
}
if(vis1[Li[i].v]==0)
{
Tarjan(Li[i].v);
low[u]=min(low[u],low[Li[i].v]);
}
if(vis2[Li[i].v])
{
Point[Num][0]=u;
Point[Num++][1]=Li[i].v;
}
}
if(low[u]==dfn[u]) //如果low[u]==dfn[u],则说明是强连通的根节点。
{
vis2[u]=true;
topo[num++] = u;
while(1)
{
if(S.empty())
{
break;
}
int v = S.top();
S.pop();
vis1[v]=2;
if(v==u)
{
break;
}
pre[v]=u;
}
}
}
void Toposort()//BFS拓扑排序
{
queue<int>Q;
for(int i=0;i<num;i++)
{
if(Du[topo[i]]==0)
{
Q.push(topo[i]);
}
}
while(!Q.empty())
{
int u=Q.front();
a[ToNum++]=u;
Q.pop();
for(int i=0;i<num;i++)
{
if(Map[u][topo[i]])
{
Du[topo[i]]--;
if(Du[topo[i]]==0)
{
Q.push(topo[i]);
}
}
}
}
}
int main()
{
int T;
int n,m;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
top = 0;
memset(Head,-1,sizeof(Head));
int u,v;
for(int i=0;i<m;i++)
{
scanf("%d %d",&u,&v);
AddEdge(u,v);
}
memset(vis1,0,sizeof(vis1));
memset(Map,0,sizeof(Map));
memset(vis2,false,sizeof(vis2));
memset(Du,0,sizeof(Du));
memset(pre,-1,sizeof(pre));
dep = 0; Num =0 ;num = 0;
while(!S.empty())
{
S.pop();
}
for(int i=1;i<=n;i++)
{
if(vis1[i]==0)
{
Tarjan(i);
}
}
for(int i=0;i<Num;i++)
{
int x = Find(Point[i][0]);
int y = Find(Point[i][1]);
Map[x][y]=1;
Du[y]++;
}
ToNum = 0;
Toposort();
bool flag=false;
for(int i=0;i<ToNum-1;i++)
{
if(!Map[a[i]][a[i+1]])//判断相邻的是不是存在边
{
flag=true;
break;
}
}
if(flag)
{
printf("No
");
}
else
{
printf("Yes
");
}
}
return 0;
}