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Description
When a directed graph is given, we can solve its transitive closure easily using the well-known Floyd algorithm.
But if you‘re given a transitive closure, can you give a corresponding directed graph with minimal edges?
Input
About 100 test cases, seperated by blank line.
First line of each case is an integer N (1<=N<=200). The following N lines represent the given transitive closure in 0-1 matrix form, each line has N numbers.
Output
For each case, just output the number of minimal edges of a directed graph which has a given transitive closure.
Sample Input
1 1 2 1 0 0 1 2 1 1 1 1 3 1 1 1 0 1 1 0 0 1
Sample Output
0 0 2 2Hint
Transitive closure can be presented as a matrix T, where Ti,j is true if and only if there is a path from vertex i to j.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
stack<int> s;
vector<int> e[205];
int n,mp[205][205],Dfs[205],low[205],use[205],isstack[205],num[205];
int top,newflag;
void init()
{
memset(mp,0,sizeof(mp));
memset(Dfs,0,sizeof(Dfs));
memset(low,0,sizeof(low));
memset(use,0,sizeof(use));
memset(isstack,0,sizeof(isstack));
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
e[i].clear();
top=0;
newflag=0;
while(!s.empty()) s.pop();
}
void tarjan(int u)
{
Dfs[u]=low[u]=++top;
s.push(u);
isstack[u]=1;
for(int i=0;i<e[u].size();i++)
{
int v=e[u][i];
if(!Dfs[v])
{
tarjan(v);
low[u]=min(low[v],low[u]);
}
else if(isstack[v])
{
low[u]=min(low[u],Dfs[v]);
}
}
if(low[u]==Dfs[u])
{
newflag++;
int x;
do
{
x=s.top();
s.pop();
use[x]=newflag;
num[newflag]++;
isstack[x]=0;
}while(x!=u);
}
}
int main()
{
int x;
while(scanf("%d",&n)!=EOF)
{
init();
int ans=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&x);
if(i!=j&&x)
e[i].push_back(j);
}
}
for(int i=1;i<=n;i++)
{
if(!Dfs[i])
tarjan(i);
}
for(int i=1;i<=newflag;i++)
{
if(num[i]!=1)
ans+=num[i];
}
for(int i=1;i<=n;i++)
{
for(int j=0;j<e[i].size();j++)
{
if(use[i]!=use[e[i][j]])
{
mp[use[i]][use[e[i][j]]]=1;
}
}
}
for(int k=1;k<=newflag;k++)
{
for(int i=1;i<=newflag;i++)
{
for(int j=1;j<=newflag;j++)
{
if(mp[i][k]&&mp[k][j]&&mp[i][j])
mp[i][j]=0;
}
}
}
for(int i=1;i<=newflag;i++)
{
for(int j=1;j<=newflag;j++)
if(mp[i][j])
ans++;
}
printf("%d\n",ans);
}
return 0;
}
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原文地址：http://www.cnblogs.com/a972290869/p/4432421.html
