(tajan+floyd) zoj 3232

It's not Floyd Algorithm

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Time Limit: 4 Seconds      Memory Limit: 32768 KB

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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
2

Hint

Transitive closure can be presented as a matrix T, where T_i,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<algorithm>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<stack>
using namespace std;
stack<int> s;
vector<int> e[205];
int n,mp[205][205],Dfs[205],low[205],cmp[205][205],use[205],num[205],top,newflag;
bool isstack[205];
void init()
{
    memset(num,0,sizeof(num));
    memset(isstack,0,sizeof(isstack));
    memset(mp,0,sizeof(mp));
    memset(cmp,0,sizeof(cmp));
    memset(low,0,sizeof(low));
    memset(use,0,sizeof(use));
    memset(Dfs,0,sizeof(Dfs));
    top=0,newflag=0;
    while(!s.empty())
        s.pop();
    for(int i=1;i<=n;i++)
        e[i].clear();
}
void tarjan(int u)
{
    Dfs[u]=low[u]=++top;
    isstack[u]=1;
    s.push(u);
    for(int i=0;i<e[u].size();i++)
    {
        int v=e[u][i];
        if(!Dfs[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(isstack[v])
            low[u]=min(low[u],Dfs[v]);
    }
    if(Dfs[u]==low[u])
    {
        newflag++;
        int x;
        do
        {
            x=s.top();
            use[x]=newflag;
            num[newflag]++;
            s.pop();
            isstack[x]=0;
        }while(x!=u);
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;
        init();
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                scanf("%d",&cmp[i][j]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(cmp[i][j]&&i!=j)
                {
                    e[i].push_back(j);
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            if(!Dfs[i])
                tarjan(i);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<e[i].size();j++)
            {
                if(use[i]!=use[e[i][j]]&&cmp[i][e[i][j]])
                {
                    mp[use[i]][use[e[i][j]]]=1;
                }
            }
        }
        for(int i=1;i<=newflag;i++)
        {
            if(num[i]!=1)
                ans+=num[i];
        }
        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;
}