题意:给出一张公交网络,求两点之间的最短路径
链接:http://acm.hdu.edu.cn/showproblem.php?pid=2112
思路:建图,floyd或者SPFA随便搞搞就过了。
注意点:无向图,并且可能存在两点之间存在多条路径,保存最短的那条即可。
以下为AC代码:
Run ID | Submit Time | Judge Status | Pro.ID | Exe.Time | Exe.Memory | Code Len. | Language | Author |
11917728 | 2014-10-19 23:37:08 | Accepted | 2112 | 1953MS | 1376K | 1997 B | G++ | luminous11 |
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define INF 0x3f3f3f3f
using namespace std;
int adj[505][505];
void floyd ( int n )
{
for ( int i = 0; i < n; i ++ )
{
for ( int j = 0; j < n; j ++ )
{
for ( int k = 0; k < n; k ++ )
{
adj[j][k] = min ( adj[j][i] +adj[i][k], adj[j][k] );
}
}
}
for ( int i = 0; i < n; i ++ )
{
adj[i][i] = 0;
}
}
int main()
{
map<string,int> p;
string be_s, en_s;
string str_1, str_2;
int len;
int n;
ios::sync_with_stdio( false );
while ( cin >> n )
{
p.clear();
memset ( adj, 0x3f3f3f3f, sizeof ( adj ) );
if ( n == -1 )
{
break;
}
int cnt = 0;
cin >> be_s >> en_s;
if ( ! p[be_s] )
{
p[be_s] = cnt ++;
}
if ( ! p[en_s] )
{
p[en_s] = cnt ++;
}
int a, b;
for ( int i = 0; i < n; i ++ )
{
cin >> str_1 >> str_2;
cin >> len;
if ( ! p[str_1] )
{
p[str_1] = cnt ++;
}
if ( ! p[str_2] )
{
p[str_2] = cnt ++;
}
if ( adj[p[str_1]][p[str_2]] > len )
adj[p[str_1]][p[str_2]] = adj[p[str_2]][p[str_1]] = len;
}
floyd ( cnt );
if ( adj[p[be_s]][p[en_s]] != INF )
{
cout << adj[p[be_s]][p[en_s]] << endl;
}
else
{
cout << "-1" << endl;
}
}
return 0;
}