COJ 1106 幻想乡的路

题意：若第二小生成树代价与最小生成树代价相等输出Not Unique!，否则输出最小代价；

对最小生成树的每一边标记后重新计算不含这条边的最小生成树，若代价与 MST 的相等说明有多种方案，复杂度为O(N^3)。

# include <stdio.h>
# include <string.h>

# define N 20005

typedef struct
{
    int u, v, w;
} Road;

char f[N], use[N];
int n, m, p[205];
Road r[N];

int cmp(const void *x, const void *y)
{
    return (*(Road*)x).w>(*(Road*)y).w ? 1:-1;
}
int find(int x)
{
    return x==p[x] ? x:(p[x] = find(p[x]));
}

int main()
{
    int i, ans, ok;

    while (~scanf("%d%d", &n, &m))
    {
        for (i = 1; i <= m; ++i)
            scanf("%d%d%d",&r[i].u, &r[i].v, &r[i].w);

        qsort(r+1, m, sizeof(Road), cmp);

        memset(f, 0, sizeof(f));
        memset(use, 0, sizeof(use));

        ok = 0;
        ans = kruskal(1);
        if (ans > 0)
        {
            for (i = 1; i <= m; ++i)
                if (use[i])
                {
                    f[i] = 1;
                    if (ans == kruskal(0))
                    {
                        ok = 1;
                        break;
                    }
                    f[i] = 0;
                }
        }
        else ok = 1;

        if (ok) puts("Not Unique!");
        else printf("%d\n", ans);
    }

    return 0;
}

int kruskal(int min)
{
    int i, ans, x, y, cnt;

    for (i = 1; i <= n; ++i)
        p[i] = i;

    cnt = ans = 0;
    for (i = 1; i <= m; ++i)
        if (!f[i])
        {
            x = find(r[i].u);
            y = find(r[i].v);
            if (x != y)
            {
                p[x] = y;
                ans += r[i].w;
                ++cnt;
                if (min) use[i] = 1;  /* 调试了很久，就是这句话写在了外面 */
            }
        }

    return cnt==n-1 ? ans:-1;
}