LightOJ-1019-Brush (V)

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題目傳送門

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題意:有N個路口和M條路，求從1路口到N路口的最短路徑。

思路:裸的最短路模板。

#include <bits/stdc++.h>

using namespace std;

int mp[120][120];
int n, m;
const int inf = 0x3f3f3f3f;
void init()
{
    for (int i = 0; i <120; i++)
    {
        for (int j = 0; j < 120; j++)
            mp[i][j] = inf;
        mp[i][i] = 0;
    }
}

int dij(int x)
{
    int dis[120], book[120];
    memset(book, 0, sizeof(book));
    for (int i = 1; i <= n; i++)
        dis[i] = mp[x][i];
    book[x] = 1;
    for (int i = 1; i <= n; i++)
    {
        int mi = inf, f= -1;
        for (int j = 1; j <= n; j++)
        {
            if (!book[j] && dis[j]<mi)
            {
                f = j;
                mi = dis[j];
            }
        }
        if (f==-1)
            break;
        book[f] = 1;
        for (int j = 1; j <= n; j++)
        {
            if (!book[j] && dis[j] > dis[f]+mp[f][j])
            {
                dis[j] = dis[f]+mp[f][j];
            }
        }
    }
    return dis[n];
}

int main(void)
{
    int T, cas=1;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d %d", &n, &m);
        init();
        while (m--)
        {
            int x, y, w;
            scanf("%d %d %d", &x, &y, &w);
            if (mp[x][y]>w)
            {
                mp[x][y] = mp[y][x] = w;
            }
        }
        int ans = dij(1);
        if (ans==inf)
            printf("Case %d: Impossible\n", cas++);
        else
            printf("Case %d: %d\n", cas++, ans);
    }
    return 0;
}