回家(洛谷 P1592)

模板题。。

传送门：codevs 1079

思路 ：以 Z 为起点 直接跑一边SPFA , 看哪一头母牛距离Z点最近 ， 最后找出Z 到 A~Y 的最短路 （因为仅有A~Z有奶牛）

#include <iostream>
#include <cstdio>
#include <cstring>
#define Max 300000
#define INF 100000000
using namespace std;
int N, Count;
int head [Max], dis [Max], queue [Max];
bool visit [Max];
struct node
{
    int to;
    int dis;
    int next;
}Edge [Max];
void AddEdge (int x, int y, int w)
{
    Count++;
    Edge [Count].to = y;
    Edge [Count].dis = w;
    Edge [Count].next = head [x];
    head [x] = Count;
}
void SPFA (int start)
{
    int head_cur = , tail_cur = ;
    for (int i = ; i <= ; i++)
    {
        dis [i] = INF;
        visit [i] = false;
    }
    dis [start] = ;
    queue [] = start;
    while (head_cur <= tail_cur)
    {
        head_cur++;
        int now = queue [head_cur];
        for (int i = head [now]; i; i = Edge [i].next)
            if (dis [now] + Edge[i].dis < dis [Edge [i].to])
            {
                dis [Edge [i].to] = dis [now] + Edge[i].dis ;
                if (visit [Edge[i].to ] == false)
                {
                    queue [++tail_cur] = Edge [i].to;
                    visit [Edge[i].to ] = true;
                }
            }
        visit [now] = false;
    }
}
int main()
{
    ios :: sync_with_stdio (false);
    cin >> N;
    char x, y;
    int w;
    for (int i = ; i <= N; i++)
    {
        cin >> x >> y >> w;
        AddEdge ((int) (x), (int) (y), w);   //直接用 字母 的Ascll 码 作为节点即可
        AddEdge ((int) (y), (int) (x), w);
    }
    SPFA ((int)'Z');
    int flag, Maxn = INF;
    for (int i = 'A'; i <= 'Y'; i++)  //因为只有A ~ Y 有母牛，所以只需查找这些点即可
    {
        if (dis [i] < Maxn)
        {
            Maxn = dis [i];
            flag = i;    //找出距离最短的点
        }
    }
    cout << (char)flag << " " << dis [flag];
    return ;
}