[次小生成树]The Unique MST

/*
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu
Submit Status Practice POJ 1679

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!

*/

# include <iostream># include <cstdio># include <algorithm>using namespace std;const int N = 105;int father[N];int m, n;struct node{    int x, y, dis, flag;}a[10005];bool cmp(node x, node y){    return x.dis < y.dis;}void init(){    for (int i=0; i<N; i++)    {        father[i] = i;    }}int find(int x){    while (x != father[x])    {        x = father[x];    }    return x;}bool Union(int x, int y){    x = find(x);    y = find(y);    if (x == y)    {        return false;    }    else    {        father[x] = y;        return true;    }}int kruskul(int num){    int ans = 0;    int cnt = 1;    for (int i=0; i<m; i++)    {        if (i == num)       //替换当前的边        {            continue;        }        if ( Union(a[i].x, a[i].y) )        {            ans += a[i].dis;            cnt++;        }    }    if (cnt != n)       //结点数要够    {        return -1;    }    else    {        return ans;    }}int main(){    int t;    int cnt, ans, sum;      //cnt 记录顶点个数  ans sum 分别记录最小生成树和次小生成树的权值之和    scanf("%d", &t);    while (t--)    {        scanf("%d %d", &n, &m);        init();        for (int i=0; i<m; i++)     //把两个点和一个权值记录在一个结构体里面        {            scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].dis);            a[i].flag = 0;        }        sort(a, a+m, cmp);              //排序        cnt = 1;        ans = 0;        for (int i=0; i<m; i++)        {            if ( Union(a[i].x, a[i].y) )            {                a[i].flag = 1;                ans += a[i].dis;                cnt++;              //统计结点数            }        }        int flag = 0;        for (int i=0; i<m; i++)        {            if (a[i].flag == 0)     //替换原来最小生成树中的边            {                continue;            }            sum = 0;            init();            sum = kruskul(i);            if (sum == ans)            {                flag = 1;                break;            }        }        if (flag)        {            printf("Not Unique!\n");        }        else        {            printf("%d\n", ans);        }    }    return 0;}