POJ 1679 The Unique MST (Kruskal 判最小生成树是否唯一)

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21646		Accepted: 7661

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!

Source
POJ Monthly--2004.06.27 srbga@POJ

题目链接：http://poj.org/problem?id=1679

题目大意：判断最小生成树是否唯一

题目分析：对图中的每条边，扫瞄其他边，如果存在权值相同的边则标记，然后用Kruskal求最小生成树，如果该MST中未包含标记的边，则必唯一直接输出权值，否则依次去掉这些边再求MST，若两次求得的权值相同，则说明不唯一

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAXN = 105;
int const MAXM = 6000;
int m, n;
int fa[MAXN], rank[MAXN];
bool fir, flag;
struct Edge
{
    int u, v, w;
    bool equal, used, del;
}e[MAXM];

bool cmp(Edge a, Edge b)  
{  
    return a.w < b.w;  
}  

void UFset()  
{  
    for(int i = 0; i <= n; i++)  
        fa[i] = i;  
    memset(rank, 0, sizeof(rank));  
}  
  
int Find(int x)  
{  
    return x == fa[x] ? x : fa[x] = Find(fa[x]);  
}  
  
void Union(int a, int b)  
{  
    int r1 = Find(a);  
    int r2 = Find(b);  
    if(r1 == r2)  
        return;  
    if(rank[r1] > rank[r2])  
        fa[r2] = r1;  
    else  
    {  
        fa[r1] = r2;  
        if(rank[r1] == rank[r2])  
            rank[r1]++;  
    }  
}  

int Kruskal()  
{  
    int u, v;  
    int num = 0, sum = 0;  
    UFset();  
    for(int i = 0; i < m; i++)  
    {  
        if(e[i].del)
            continue;
        u = e[i].u;  
        v = e[i].v;  
        if(Find(u) != Find(v))  
        {  
            sum += e[i].w;  
            Union(u, v);  
            num++;  
            if(fir)
                e[i].used = true;
        }  
        if(num == n - 1)  
            break;  
    }  
    return sum;
} 

int main()
{
    int T, u, v, w;
    scanf("%d", &T);
    while(T--)
    {
        flag = false;
        scanf("%d %d", &n, &m);
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
            e[i].equal = e[i].used = e[i].del = false;
        }
        sort(e, e + m, cmp);
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(i == j)
                    continue;
                if(e[i].w == e[j].w)
                    e[i].equal = true;
            }
        }
        fir = true;
        int w1 = Kruskal(), w2;
        fir = false;
        for(int i = 0; i < m; i++)
        {
            if(e[i].used && e[i].equal)
            {
                e[i].del = true;    //删去这条边
                w2 = Kruskal();
                if(w1 == w2)
                {
                    printf("Not Unique!\n");
                    flag = true;
                    break;
                }
                e[i].del = false;
            }
        }
        if(!flag)
            printf("%d\n", w1);
    }
}