POJ_1679_The Unique MST(次小生成树模板)

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23942		Accepted: 8492

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!

题意：问最小生成树是否唯一。

分析：求次小生成树，推断次小生成树和最小生成树是否相等。留作模板。

次小生成树的步骤：

(1)先用Prime求出最小生成树T，在Prime的同一时候用一个矩阵max_edge[u][v]记录在T中连接随意两点u,v的唯一路径中权

值最大的那条边的权值。注意这里是非常easy做到的。由于Prime是每次添加一个节点t。而设已经标了号的节点集合为S，则S

中全部节点到t的路径中最大权值的边就是当前增加的这条边。

(2)枚举最小生成树以外的边，并删除该边所在环上权值最大的边。

(3)取得的全部生成树中权值最小的一棵即为所求。

算法的时间复杂度为O(n^2)。

题目链接：

id=1679">http://poj.org/problem?id=1679

代码清单：

#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<string>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 100 + 5;
const int maxv = 10000 + 5;
const int max_dis = 1e9 + 5;

int T;
int n,m;
int a,b,c;
int MST,_MST;
bool vis[maxn];
int father[maxn];
int dist[maxn];
int graph[maxn][maxn];
bool used[maxn][maxn];
int max_edge[maxn][maxn];

void init(){
    memset(vis,false,sizeof(vis));
    memset(used,false,sizeof(used));
    memset(max_edge,-1,sizeof(max_edge));
    memset(graph,0x7f,sizeof(graph));
}

void input(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++){
        scanf("%d%d%d",&a,&b,&c);
        graph[a][b]=graph[b][a]=c;
        used[a][b]=used[b][a]=true;
    }
}

int prim(){
    int ans=0;
    dist[1]=0;
    vis[1]=true;
    father[1]=-1;
    for(int i=2;i<=n;i++){
        father[i]=1;
        dist[i]=graph[1][i];
    }
    for(int i=1;i<n;i++){
        int v=-1;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&(v==-1||dist[j]<dist[v])) v=j;
        }
        ans+=dist[v];
        vis[v]=true;
        used[father[v]][v]=used[v][father[v]]=false;
        for(int j=1;j<=n;j++){
            if(vis[j]){
                max_edge[v][j]=max_edge[j][v]=max(max_edge[father[v]][j],dist[v]);
            }
            else{
                if(graph[v][j]<dist[j]){
                    dist[j]=graph[v][j];
                    father[j]=v;
                }
            }
        }
    }return ans;
}

int second_prim(){
    int ans=max_dis;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(used[i][j]) ans=min(ans,MST+graph[i][j]-max_edge[i][j]);
    return ans;
}

void solve(){
    MST=prim();
    _MST=second_prim();
    if(MST==_MST) printf("Not Unique!\n");
    else printf("%d\n",MST);
}

int main(){
    scanf("%d",&T);
    while(T--){
        init();
        input();
        solve();
    }return 0;
}