POJ 1679 The Unique MST

求出次小生成树与最小生成树比较，如果相等则不唯一，否则输出最小生成树的权值。

这道题整整花了两个小时的时间，用的是prim算法，用Tlink记录边是否在MST中，然
后枚举不在树中的边，用其权值w[u][v]代替max[u][v]。max[u][v]是MST中连结两

点u,v唯一的路中权值最大的那条边的权值。

/*
Memory: 272K  Time: 0MS
Language: C++  Result: Accepted
*/


#include<cstdio>
#include<cstring>
#include<cstdlib>
const int inf = 0x3f3f3f3f;
const int MAXN = 105;
bool Tlink[MAXN][MAXN], vis[MAXN];
int w[MAXN][MAXN], lowc[MAXN], pre[MAXN], max[MAXN][MAXN];
int n, m;

int Max( int a, int b)
{
    return a > b ? a : b;
}

int prim()
{
    int i, j, p, k;
    int minc, res = 0;
    memset( vis, false, sizeof vis);
    memset( pre, 0, sizeof pre);
    memset( max, 0, sizeof max);
    vis[1] = 1;
    pre[1] = 1;
    for( i = 2; i <= n; i ++)  { lowc[i] = w[1][i]; pre[i] = 1;}
    for( i = 2; i <= n; i ++)
    {
        minc = inf; p = -1;
        for( j = 1; j <= n; j ++) {
            if( !vis[j] && lowc[j] < minc){
                minc = lowc[j];
                p = j;
            }
        }
        vis[p] = true;
        res += minc;
        max[ pre[p] ][p] = minc;
        Tlink[ pre[p] ][p] = true;
        Tlink[p][ pre[p] ] = true;
        for( k = 1; k <= n; k ++)
            max[k][p] = Max( max[ pre[p] ][p], max[k][p]);
        for( j = 1; j <= n; j ++)
            if( !vis[j] && lowc[j] > w[p][j]) {
                lowc[j] = w[p][j];
                pre[j] = p;
            }
    }
    return res;
}

int main()
{
    int T;
    scanf( "%d", &T);
    int s, e, t, Ans, ans;
    while( T --)
    {
        scanf( "%d%d", &n, &m);
        for( int i = 1; i <= n; i ++)
            for( int j = 1; j <= n; j ++)
                w[i][j] = inf;
        memset( Tlink, false, sizeof Tlink);
        for( int i = 1; i <= m; i ++)
        {
            scanf( "%d%d%d", &s, &e, &t);
            w[s][e] = t;
            w[e][s] = t;
        }
        bool ok = true;
        Ans = prim();
        for( int i = 1; i <= n; i ++)
        {
            for( int j = 1; j <= n; j ++)
            {
                if( w[i][j] == inf || Tlink[i][j])
                    continue;
                ans = Ans + w[i][j] - max[i][j];
                if( Ans == ans) {
                    ok = false;
                    break;
                }
            }
            if( !ok) break;
        }
        if( ok) printf( "%d\n", Ans);
        else printf( "Not Unique!\n");
    }
    return 0;
}