hdu 3836 Equivalent Sets（強連通分量--加邊）
阿新 • • 發佈：2017-08-06
accep nod ons first pan val while 無環 mono
Total Submission(s): 2798 Accepted Submission(s): 962
Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
Input The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
Output For each case, output a single integer: the minimum steps needed.
Sample Input
Sample Output
Equivalent Sets
Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 104857/104857 K (Java/Others)Total Submission(s): 2798 Accepted Submission(s): 962
Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
Input The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
Output For each case, output a single integer: the minimum steps needed.
Sample Input
4 0 3 2 1 2 1 3
Sample Output
4
2
HintCase 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
求將原圖的強連通分量縮點，得到有向無環圖，求至少加多少條邊能夠使這個圖變成一幅強連通圖，max（入度為0的點，出度為0的點）即為答案。
#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"algorithm"
using namespace std;
#define N 20005
#define M 50005
#define min(a,b) (a<b?a:b)
const int inf=1000000;
struct node
{
int u,v,next;
}e[M];
int t,bcnt,index,stop,ans;
int head[N],dfn[N],low[N],stap[N],mark[N],be[N];
int indeg[N],out[N];
void add(int u,int v)
{
e[t].u=u;
e[t].v=v;
e[t].next=head[u];
head[u]=t++;
}
void tarjan(int u)
{
int i,v;
dfn[u]=low[u]=++index;
stap[++stop]=u;
mark[u]=1;
for(i=head[u];i!=-1;i=e[i].next)
{
v=e[i].v;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(mark[v])
low[u]=min(low[u],dfn[v]);
}
if(low[u]==dfn[u])
{
bcnt++;
do
{
v=stap[stop--];
mark[v]=0;
be[v]=bcnt;
}
while(u!=v);
ans++;
}
}
void solve(int n)
{
int i;
memset(dfn,0,sizeof(dfn));
index=stop=bcnt=0;
for(i=1;i<=n;i++)
{
if(!dfn[i])
tarjan(i);
}
}
void work(int n)
{
int i,j,u,v,t1,t2;
memset(indeg,0,sizeof(indeg));
memset(out,0,sizeof(out));
for(i=1;i<=n;i++)
{
u=be[i];
for(j=head[i];j!=-1;j=e[j].next)
{
v=be[e[j].v];
if(u!=v)
{
indeg[v]++;
out[u]++;
}
}
}
t1=t2=0;
for(i=1;i<=bcnt;i++)
{
if(indeg[i]==0)
t1++;
if(out[i]==0)
t2++;
}
printf("%d\n",t1>t2?t1:t2);
}
int main()
{
int n,m,u,v;
while(scanf("%d%d",&n,&m)!=-1)
{
t=0;
memset(head,-1,sizeof(head));
while(m--)
{
scanf("%d%d",&u,&v);
add(u,v);
}
ans=0;
solve(n);
// printf("%d\n",ans);
if(ans==1)
printf("0\n");
else
work(n);
}
return 0;
}hdu 3836 Equivalent Sets（強連通分量--加邊）