【USACO2.4.4】回家 最短路

想说：什么破题目描述。

自环！？  从A -> a 距离是100的话，a -> C 距离是50. 那么 那么A->C距离是150 ！

题目需要把A和a看成是2个截然不同的点！然后直接做最短路即可。  这题目描述我看的也是醉了……

   Test 1: TEST OK [0.003 secs, 3404 KB]
   Test 2: TEST OK [0.003 secs, 3404 KB]
   Test 3: TEST OK [0.005 secs, 3404 KB]
   Test 4: TEST OK [0.003 secs, 3404 KB]
   Test 5: TEST OK [0.003 secs, 3404 KB]
   Test 6: TEST OK [0.005 secs, 3404 KB]
   Test 7: TEST OK [0.016 secs, 3404 KB]
   Test 8: TEST OK [0.003 secs, 3404 KB]
   Test 9: TEST OK [0.003 secs, 3404 KB]

All tests OK.

/*
TASK:comehome
LANG:C++
*/

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

int n;
int f[100][100];
bool cow[26]={0};

int check(char ch)
{
	if ('A' <= ch && ch <= 'Z')	return ch -  'A' + 26;
	else return ch - 'a';
}

void init()
{
	int n, tmp;
	scanf("%d",&n);
	memset(f, 50, sizeof(f));
	while (n--)
	{
		getchar();
		int X = check(getchar());
		getchar();
		int Y = check(getchar());	
		scanf("%d", &tmp);
		//cout<<X<<" "<<Y<<" "<<tmp<<endl;
		f[X][Y] = f[Y][X] = min(f[X][Y], tmp);
	}
}

void doit()
{
	for (int k = 0; k != 52; ++ k)
		for (int i = 0; i != 52; ++ i)
			for (int j =0; j != 52; ++ j)
				if (f[i][j] > f[i][k] + f[k][j])	f[i][j] = f[j][i] = f[i][k] + f[k][j];
	int wz, ans=0x7fffffff;
	for (int i = 26; i != 51; ++ i)
		if (f[51][i] < ans)
		{
			ans = f[51][i];
			wz = i - 26;	
		}
	cout<< char(wz + 'A') <<" "<<ans<<endl;
}

int main()
{
	freopen("comehome.in","r",stdin);
	freopen("comehome.out","w",stdout);
	init();
	doit();
	return 0;
}