POJ -1679（次小生成树）模板

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions:34617		Accepted: 12637

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

Sample Output

3

Not Unique!

题意：求是否有第二个最小生成树
（次小生成树模板）

#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 =  + ;

const int maxv =  + ;

const int max_dis = 1e9 + ;

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,-,sizeof(max_edge));

    memset(graph,0x7f,sizeof(graph));

}

void input() {

    scanf("%d%d",&n,&m);

    for(int i=; 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=;

    dist[]=;

    vis[]=true;

    father[]=-;

    for(int i=; i<=n; i++) {

        father[i]=;

        dist[i]=graph[][i];

    }

    for(int i=; i<n; i++) {

        int v=-;

        for(int j=; j<=n; j++) {

            if(!vis[j]&&(v==-||dist[j]<dist[v])) v=j;

        }

        ans+=dist[v];

        vis[v]=true;

        used[father[v]][v]=used[v][father[v]]=false;

        for(int j=; 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=; i<=n; i++)

        for(int j=; 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 ;

}

巴特西

POJ -1679（次小生成树）模板

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