CodeForces

http://codeforces.com/problemset/problem/601/A

这道题没想过来， 有点脑筋急转弯的感觉了

本质上就是找最短路径 但是卡在不能重复走同一个点 ---->>> 这是来坑人的

因为这是一个完全图(不是被road 连接  就是被rail连接 ) 所以一定有一条直接连1 和 n的路径

那么只用找没有连 1 和 n 的路径的 那个图的最短路即可

然后这个dijkstra写的是O(V^2)的写法 

以后尽量用优先队列的写法O(ElogV)

#include <iostream>
#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
using namespace std;

int rail[401][401];
int road[401][401];
int n, m;

int dijkstra(int town[401][401])
{
    int dist[401];
    bool use[401];
    memset(use, 0, sizeof(use));
    fill(dist, dist+n+1, INF);
    dist[1] = 0;
    while (true)
    {
        int v = -1;
        for (int i = 1; i <= n; i++)
        {
            if (!use[i] && (v == -1 || dist[i] < dist[v])) v = i;
        }
        if (v == -1) break;
        use[v] = true;
        for (int i = 1; i <= n; i++)
        {
            dist[i] = min(dist[i], dist[v] + town[v][i]);
        }
    }
    return dist[n];

}
int main()
{
    freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            rail[i][j] = INF;
            road[i][j] = 1;
        }
    }
    for (int i = 0; i < m; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        rail[x][y] = 1;
        rail[y][x] = 1;
        road[x][y] = INF;
        road[y][x] = INF;
    }
    int ans = 0;
    if (rail[1][n] == INF || rail[n][1] == INF) ans = dijkstra(rail);
    else ans = dijkstra(road);
    if (ans == INF) printf("-1
");
    else printf("%d
",ans);
}