ZJU 1589 Professor John

分析：经典的传递闭包，用floyd-warshall算法，是O(n^3)。

for(int k = 1; k <= n; k++)

          for(int i = 1; i <= n; i++)

              for(int j = 1; j <= n; j++)

 

Code
#include <iostream>
using namespace std;

bool Map[26][26];
bool LongLink[26][26];
void Floyd_Warshall()
{
    memcpy(LongLink,Map,sizeof(Map));
    
    for(int k=0;k<26;++k)
        for(int i=0;i<26;++i)
            for(int j=0;j<26;++j)
                LongLink[i][j] |= (LongLink[i][k] && LongLink[k][j]);
}
int main()
{
    int t,i=1;
    for(cin>>t;i!=t+1;++i)
    {
        int n;
        char word[4];
        memset(Map,0,sizeof(Map));
        
        cin>>n;
        for(int j=0;j<n;++j)
        {
            cin>>word;
            if(word[1]=='>')
                swap(word[0],word[2]);
            Map[word[0]-'A'][word[2]-'A'] = true;
        }
        
        Floyd_Warshall();
        
        bool flag = false;
        cout<<"Case "<<i<<":"<<endl;
        for(int j=0;j<26;++j)
            for(int k=0;k<26;++k)
                if(j!=k && !Map[j][k] && LongLink[j][k])
                {
                    cout<<(char)(j+'A')<<"<"<<(char)(k+'A')<<endl;
                    flag = true;
                }
        if(!flag)
            cout<<"NONE"<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/jaskist/archive/2009/05/09/1453184.html