poj1679(prim或者krusal的变形）

The Unique MST

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 29778 Accepted: 10660

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!

Source

POJ Monthly--2004.06.27 srbga@POJ

 判断最小生成树的唯一性 用删边来判断  具体看注释

#include <bits/stdc++.h>#define mod 1000000007#define inf 0x3f3f3f3f#define pi acos(-1.0)using namespace std;typedef long long ll;const int N=11000;const int M=15005;int n,m,cnt;int parent[N];bool flag;struct man {    int u,v,w;    int eq,used,del;} edg[N];bool cmp(man g,man h) {return g.w<h.w;}void init() {    for(int i=0; i<=10005; i++) {        parent[i]=i;    }}int Find(int x) {    if(parent[x] != x) parent[x] = Find(parent[x]);    return parent[x];}//查找并返回节点x所属集合的根节点void Union(int x,int y) {    x = Find(x);    y = Find(y);    if(x == y) return;    parent[y] = x;}//将两个不同集合的元素进行合并int Kruskal() {    init();   int sum=0;   int num=0;   for(int i=0;i<m;i++){    if(edg[i].del==1)continue;    int u=edg[i].u;int v=edg[i].v;int w=edg[i].w;    if(Find(u)!=Find(v)){        sum+=w;        if(!flag)edg[i].used=1;//标记这条边有纳入生成树中         num++;        Union(u,v);    }    if(num>=n-1)break;   }   return sum;}int main() {    int t,d;    cin>>t;    while(t--) {        cnt=0;        cin>>n>>m;        for(int i=0; i<m; i++) {            cin>>edg[i].u>>edg[i].v>>edg[i].w;            edg[i].del=0;            edg[i].used=0;            edg[i].eq=0;//一开始这个地方eq没有初始化，WA了好几发，操        }        for(int i=0;i<m;i++){            for(int j=0;j<m;j++){                if(i==j)continue;                if(edg[i].w==edg[j].w)edg[i].eq=1;//把有相等的边标记             }        }        sort(edg,edg+m,cmp);        flag=false;        cnt=Kruskal();        flag=true;        bool gg=false;        for(int i=0;i<m;i++){            if(edg[i].used==1&&edg[i].eq==1){ //判断 如果这条边有在生成树中且有其他边与之相等                 edg[i].del=1;//删除这条边                 int s=Kruskal();//printf("%d %d\n",i,s);                if(s==cnt){                    gg=true;                    printf("Not Unique!\n");                    break;                }                edg[i].del=0; //取消删除标记             }        }        if(!gg)cout<<cnt<<endl;    }    return 0;}