POJ 1679 The Unique MST 次小生成树

题目描述：

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

23 31 2 12 3 23 1 34 41 2 22 3 23 4 24 1 2

Sample Output

3Not Unique!

题目分析：

给一个图，求这个图的最小生成树是否唯一。
其实就是找次小生成树的权值之和与最小生成树是否相等。
用Kruskal将每一条最小生成树上的边删除，在求一次kruskal。

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=110;const int maxm=10010;struct Edge{    int u,v,w;    bool f;} edge[maxm];int tol;int f[maxn];int n,m,T,ans;void addedge(int u,int v,int w){    edge[tol].u=u;    edge[tol].v=v;    edge[tol].w=w;    edge[tol++].f=0;}bool cmp(Edge a,Edge b){    return a.w<b.w;}int find(int x){    if (f[x]==-1) return x;    else return f[x]=find(f[x]);}int kruskal(){    int ret=0;    int cnt=0;    for(int i=0; i<m; i++)    {        int u=edge[i].u;        int v=edge[i].v;        int w=edge[i].w;        int t1=find(u);        int t2=find(v);        if (t1!=t2)        {            edge[i].f=1;            ret+=w;            cnt++;            f[t1]=t2;        }        if (cnt==n-1) break;    }    if (cnt<n-1) return -1;    else return ret;}int kruskal1(int x){    int ret=0;    int cnt=0;    for(int i=0; i<m; i++)    {        if (i==x) continue;        int u=edge[i].u;        int v=edge[i].v;        int w=edge[i].w;        int t1=find(u);        int t2=find(v);        if (t1!=t2)        {            ret+=w;            cnt++;            f[t1]=t2;        }        if (cnt==n-1) break;    }    if (cnt<n-1) return -1;    else return ret;}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        memset(f,-1,sizeof(f));        tol=0;        for(int i=0; i<m; i++)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w);        }        sort(edge,edge+m,cmp);        ans=kruskal();        bool flag=true;        for(int i=0; i<m; i++)        {            if (!edge[i].f) continue;            int sum=0;            memset(f,-1,sizeof(f));            sum=kruskal1(i);            if (sum==ans)            {                flag=false;                break;            }        }        if (flag) printf("%d\n",ans);        else printf("Not Unique!\n");    }    return 0;}