UVa 10986 Dijkstra TLE

时限是 3s， 当然不能用 dijksra；

N 最大是 20005，数组 d[N][N] 可以开；

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# include <stdio.h>

# include <string.h>



# define N 20005

# define M 50005

# define INF N*M



int n, m, s, t;

int w[N][N];



int min(int x, int y)

{

    return (x<y ? x:y);

}



void init(void)

{

    int i, j, k, c;

    

    scanf("%d%d%d%d", &n, &m, &s, &t);

    for (i = 0; i < n; ++i)

    for (j = 0; j < n; ++j)

    {

        w[i][j] = INF;

    }

    for (k = 1; k <= m; ++k)

    {

        scanf("%d%d%d", &i, &j, &c);

        w[i][j] = w[j][i] = c;

    }

}



void solve(void)

{

    char v[N];

    int d[N], m, x, i, y;

    

    memset(v, 0, sizeof(v));

    for (i = 0; i < n; ++i) d[i] = (i==s ? 0:INF);

    for (i = 0; i < n; ++i)

    {

        m = INF;

        for (y = 0; y < n; ++y) if (!v[y] && d[y]<m) m = d[x = y];

        //if (x == t) break;

        v[x] = 1;

        for (y = 0; y < n; ++y) d[y] = min(d[y], d[x]+w[x][y]);

    }

    if (d[t] >= INF) printf("unreachable\n");

    else printf("%d\n", d[t]);

}



int main()

{

    int T, i;

        

    scanf("%d", &T);

    for (i = 1; i <= T; ++i)

    {

        printf("Case #%d: ", i);

        init();

        solve();

    }

    

    return 0;

}

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