POJ 1679 The Unique MST

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique. 

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V. 
2. T is connected and acyclic. 

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'. 

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output

3
Not Unique!

判断最小生成树是不是唯一的，那么只需要判断最小生成树与次小生成树是否相等，若相等则不是唯一的，若不等则是唯一的：

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
#define N 110
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
int G[N][N], M[N][N], dist[N], f[N], n;
int visit[N], mark[N][N]; //标记数组
void Init()
{
    int i, j;
    memset(visit, 0, sizeof(visit));
    memset(mark, 0, sizeof(mark));
    memset(M, 0, sizeof(M));
    for (i = 0; i <= n; i++)
    {
        for (j = 0; j <= n; j++)
        {
            if (i == j) G[i][j] = 0;
            else G[i][j] = INF;
        }
    }
}
int Prim()
{
    int i, j, Min, idex, ans = 0;
    visit[1] = 1;
    for (i = 1; i <= n; i++)
    {
        dist[i] = G[1][i];
        f[i] = 1;
    }
    for (i = 1; i < n; i++)
    {
        Min = INF;
        for (j = 1; j <= n; j++)
        {
            if (!visit[j] && dist[j] < Min)
            {
                Min = dist[j];
                idex = j;
            }
        }
        ans += Min;
        visit[idex] = 1; //将已经在最小生成树的结点标记下来
        mark[f[idex]][idex] = mark[idex][f[idex]] = 1; //将已经在最小生成树里的边标记下来
        for (j = 1; j <= n; j++)
        {
            if (visit[j] && j != idex)
                M[j][idex] = M[idex][j] = max(M[f[idex]][j], G[f[idex]][idex]); //记录两个最小生成树结点之间最大的一条边
                //M[j][idex] = M[idex][j] = max(M[f[idex]][j], dist[idex]);
            if (!visit[j] && dist[j] > G[idex][j])
            {
                dist[j] = G[idex][j];
                f[j] = idex; //更新上一个结点
            }
        }
    }
    return ans;
}
int Second(int ans)
{
    int i, j, Min = INF;
    for (i = 1; i <= n; i++)
    {
        for (j = i+1; j <= n; j++)
        {
            if (!mark[i][j] && G[i][j] != INF)
                Min = min(Min, ans+G[i][j]-M[i][j]); //每次遍历时加上一个不是最小生成树中的边，需要将形成回路的最小生成树最大边减去才能形成次小生成树
        }
    }
    return Min;
}
int main ()
{
    int T, m, a, b, c, ans1, ans2;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d %d", &n, &m);
        Init();
        while (m--)
        {
            scanf("%d %d %d", &a, &b, &c);
            G[a][b] = c;
            G[b][a] = c;
        }
        ans1 = Prim(); //计算最小生成树的权值
        ans2 = Second(ans1); //计算次小生成树的权值
        if (ans1 == ans2) printf("Not Unique!
");
        else printf("%d
", ans1);
    }
    return 0;
}