POJ 1679：The Unique MST（次小生成树&&Kruskal）

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19941		Accepted: 6999

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!

这道题就是要你推断是否为唯一的最小生成树。。

这也是我第一道次小生成树的题。。那个PDF资料真是太好了。。我用的是Kruskal。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>

using namespace std;

const int max1 = 105;
const int max2 = 10005;

struct node
{
    int u, v, w;//边的顶点以及权值
    int r;//用于记录边的序号
    int used;//用于记录边是否被使用过
}edge[max2];//边的数组

int parent[max1];//顶点i所在集合相应树中的根节点
int n, m;//顶点数，边的个数
int cas;

void start()//初始化
{
    for(int i=1; i<=n; i++)
        parent[i] = i;
}

int find (int x)//查找并返回节点x所属集合的根节点
{
    return parent[x] == x ?
 x : parent[x] = find ( parent[x] );
}

bool cmp ( node a, node b )//按权值从小到大排序
{
    return a.w < b.w;
}

int Kruskal()//第一次求最小生成树
{
    sort( edge+1, edge+m+1, cmp );
    int ans = 0;
    for(int i=1; i<=m; i++)
    {
        int e = edge[i].r;
        int x = find( edge[i].u );
        int y = find( edge[i].v );
        if( x!=y )
        {
            parent[y] = x;
            ans += edge[i].w;
            edge[e].used = 1;
        }
    }
    return ans;
}

int Kruskal_again(int tt)//求次小生成树
{
     sort( edge+1, edge+m+1, cmp );
    int ans = 0;
    for(int i=1; i<=m; i++)
    {
        if( tt==edge[i].r )
            continue;
        int x = find( edge[i].u );
        int y = find( edge[i].v );
        if( x!=y )
        {
            parent[y] = x;
            ans += edge[i].w;
        }
    }
    return ans;
}

bool judge()//推断是否能就得最小生成树，即看是否有孤立的点
{
    for(int i=1; i<=n; i++)
        if( find(i) != find(1) )
        return false;
    return true;
}

int main()
{
    scanf("%d", &cas);
    while( cas-- )
    {
        scanf("%d%d", &n, &m);
        start();
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
            edge[i].used = 0;
            edge[i].r = i;
        }
        int flag = 0;
        int ans1 = Kruskal();
        for(int i=1; i<=m; i++)//一一枚举求最小生成树。。
        {
            if( !edge[i].used )
                continue;
            start();
            int ans2 = Kruskal_again(i);//求除去此边的最小生成树
            if( ans1 == ans2 && judge() )
            {
                flag = 1;//标记结论
                break;
            }
        }
        if( flag )
            printf("Not Unique!
");
        else
            printf("%d
", ans1);
    }
    return 0;
}