poj_1679_判断最小生成树是否唯一

The Unique MST

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!
mean
第一行输入t,代表t组案例,下一行输入n,m
代表有n个村,m 条边
接下来m行输入 u,v,w u ,v村的位置,w权值
ans
什么情况下,最小生成树不唯一,在有权值相同的边的情况下,枚举每个边

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;int n, m;struct node{    int u, v, w;    int used, equ, del;} nn[99999];int pre[99999], flag;void init(){    for (int i = 0; i <= n; i++)        pre[i] = i;}bool cmp(node a, node b){    return a.w<b.w;}int Find(int x){    return pre[x] == x ? x : Find(pre[x]);}int solve(){    int i, j, ans = 0, cnt = 0;    init();    for (i = 0; i<m; i++)    {        if (nn[i].del) continue;        int fx = Find(nn[i].u), fy = Find(nn[i].v);        if (fx != fy)        {            if (!flag)                nn[i].used = 1;            ans += nn[i].w;            cnt++;            pre[fx] = fy;        }        if (cnt == n - 1)            break;    }    if (cnt != n - 1)        ans = -1;    return ans;}int main(){    int t, i, j;    cin >> t;    {        while (t--)        {            cin >> n >> m;            for (i = 0; i<m; i++)            {                scanf("%d%d%d", &nn[i].u, &nn[i].v, &nn[i].w);                nn[i].del = nn[i].equ = nn[i].used = 0;            }            sort(nn, nn + m, cmp);            for (i = 0; i<m; i++)            {                for (j = 0; j<m; j++)                {                    if (i != j&&nn[i].w == nn[j].w)                        nn[i].equ = nn[j].equ = 1;                }            }             flag = 0;            int ans = solve();            // puts("(*(*");            //  cout<<ans<<endl;            flag = 1;            if (ans == -1)            {                puts("Not Unique!");                continue;            }            int f = 0;            for (i = 0; i<m; i++)            {                if (nn[i].equ&&nn[i].used)                {                    nn[i].del = 1;                    int tmp = solve();                    if (tmp == ans)                    {                        f = 1;                        puts("Not Unique!");                        break;                    }                    nn[i].del = 0;                }            }            if (!f)                cout << ans << endl;        }    }}