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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