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Going from u to v or from v to u?
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 14562 | Accepted: 3844 |
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
一开始想成只要判断强连通分量的个数是否是1就可以了,发现怎么也不对，再仔细读了题，发现对于x和y，只有有一方可以到另一方就可以了，所以做法是tarjan缩点然后找入度为0的点和出度为0的点是否有多个，如果入度为0的点超过1个或者出度为0的点超过1个就不行，
如图：
第一张图是入度为0的点有2个，显然上面两个点不满足题意，第二张图是出度为0的点有2个，显然上面的两个点不满足题意
#include<map>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1010;
const int M = 6100;
const int inf = 0x3f3f3f3f;
int DFN[N];
int low[N];
int block[N];
int Stack[N];
int in[N];
int out[N];
bool instack[N];
int head[N];
int cost[N];
int min_price[N];
int tot, sccnum, index, top, n, m;
struct node
{
int next;
int to;
}edge[M];
void addedge(int from, int to)
{
edge[tot].to = to;
edge[tot].next = head[from];
head[from] = tot++;
}
void tarjan(int u)
{
DFN[u] = low[u] = ++index;
Stack[top++] = u;
instack[u] = 1;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (DFN[v] == 0)
{
tarjan(v);
if (low[u] > low[v])
{
low[u] = low[v];
}
}
else if (instack[v])
{
if (low[u] > DFN[v])
{
low[u] = DFN[v];
}
}
}
if (DFN[u] == low[u])
{
sccnum++;
do
{
top--;
block[Stack[top]] = sccnum;
instack[Stack[top]] = 0;
}while ( top >=0 && Stack[top] != u);
}
}
void solve()
{
memset( instack, 0, sizeof(instack) );
memset( DFN, 0, sizeof(DFN) );
memset( low, 0, sizeof(low) );
memset( in, 0, sizeof(in) );
memset( out, 0, sizeof(out) );
memset( min_price, inf, sizeof(min_price) );
sccnum = index = top = 0;
for (int i = 1; i <= n; i++)
{
if (DFN[i] == 0)
{
tarjan(i);
}
}
for (int i = 1; i <= n; i++)
{
for (int j = head[i]; j != -1; j = edge[j].next)
{
if (block[i] != block[edge[j].to])
{
out[block[i]]++;
in[block[edge[j].to]]++;
}
}
}
int zero_in = 0;
int zero_out = 0;
for (int i = 1; i <= sccnum; i++)
{
if (in[i] == 0)
{
zero_in++;
}
if (out[i] == 0)
{
zero_out++;
}
}
if(zero_in >= 2 || zero_out >= 2)
{
printf("No\n");
}
else
{
printf("Yes\n");
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
memset( head, -1, sizeof(head) );
tot = 0;
int u, v;
for (int i = 0; i < m; i++)
{
scanf("%d%d", &u, &v);
addedge(u, v);
}
solve();
}
return 0;
}
#include<map>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1010;
const int M = 6100;
const int inf = 0x3f3f3f3f;
int DFN[N];
int low[N];
int block[N];
int Stack[N];
int in[N];
int out[N];
bool instack[N];
int head[N];
int cost[N];
int min_price[N];
int tot, sccnum, index, top, n, m;
struct node
{
int next;
int to;
}edge[M];
void addedge(int from, int to)
{
edge[tot].to = to;
edge[tot].next = head[from];
head[from] = tot++;
}
void tarjan(int u)
{
DFN[u] = low[u] = ++index;
Stack[top++] = u;
instack[u] = 1;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (DFN[v] == 0)
{
tarjan(v);
if (low[u] > low[v])
{
low[u] = low[v];
}
}
else if (instack[v])
{
if (low[u] > DFN[v])
{
low[u] = DFN[v];
}
}
}
if (DFN[u] == low[u])
{
sccnum++;
do
{
top--;
block[Stack[top]] = sccnum;
instack[Stack[top]] = 0;
}while ( top >=0 && Stack[top] != u);
}
}
void solve()
{
memset( instack, 0, sizeof(instack) );
memset( DFN, 0, sizeof(DFN) );
memset( low, 0, sizeof(low) );
memset( in, 0, sizeof(in) );
memset( out, 0, sizeof(out) );
memset( min_price, inf, sizeof(min_price) );
sccnum = index = top = 0;
for (int i = 1; i <= n; i++)
{
if (DFN[i] == 0)
{
tarjan(i);
}
}
for (int i = 1; i <= n; i++)
{
for (int j = head[i]; j != -1; j = edge[j].next)
{
if (block[i] != block[edge[j].to])
{
out[block[i]]++;
in[block[edge[j].to]]++;
}
}
}
int zero_in = 0;
int zero_out = 0;
for (int i = 1; i <= sccnum; i++)
{
if (in[i] == 0)
{
zero_in++;
}
if (out[i] == 0)
{
zero_out++;
}
}
if(zero_in >= 2 || zero_out >= 2)
{
printf("No\n");
}
else
{
printf("Yes\n");
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
memset( head, -1, sizeof(head) );
tot = 0;
int u, v;
for (int i = 0; i < m; i++)
{
scanf("%d%d", &u, &v);
addedge(u, v);
}
solve();
}
return 0;
}
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