POJ-1679-The Unique MST（次小生成树）

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The Unique MST

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 26364 Accepted: 9419

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

题意：给定一棵树，若有多个MST或无MST则输出Not Unique，否则输出MST大小。

在寻找最小生成树过程中，标记出MST边与非MST边，并求出MST中两点（u，v）之间的最大边MAX[U][V] ，

设非MST中边的集合为P,非MST中边的集合为T；

则寻找其次小生成树只需寻找 用P的边uv替换掉T中的MAX[u][v]后的最小值  即 min（ans+cost[u][v]-MAX[u][v]），

而此题中只需判断是否存在P中边uv等于MAX[u][v]（替换边等于最大边）即可，

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <cstring>#include <algorithm>#include <string>#include <set>#include <stack>#include <map>#include <queue>#define INF 0x3f3f3f3fusing namespace std;const int MAXN = 105;int cost[MAXN][MAXN];bool vis[MAXN];bool in_MST[MAXN][MAXN];int Max_e[MAXN][MAXN];int lowcost[MAXN];int pre[MAXN];int m,n;int prim(){    int ans=0;    memset(vis,0,sizeof(vis));    memset(Max_e,0,sizeof(Max_e));    memset(in_MST,0,sizeof(in_MST));    vis[0]=1;    pre[0]=-1;    for(int i=0; i<n; ++i)    {        lowcost[i]=cost[0][i];        pre[i]=0;    }    lowcost[0]=0;    for(int i=1; i<n; ++i)    {        int minc=INF;        int p=-1;        for(int j=0; j<n; ++j)            if(!vis[j] && minc>lowcost[j])        {            minc=lowcost[j];            p=j;        }        if(minc==INF)return -1;        ans+=minc;        vis[p]=true;        in_MST[p][ pre[p] ]=in_MST[ pre[p] ][p]=true;        for(int j=0; j<n; ++j)        {            if(vis[j])            Max_e[j][p]=Max_e[p][j]=max(Max_e[ pre[p] ][j],lowcost[p]);            if(!vis[j]&&lowcost[j]>cost[p][j])            {                lowcost[j]=cost[p][j];                pre[j]=p;            }        }    }    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int x,y,c;        memset(cost,INF,sizeof(cost));        scanf("%d%d",&n,&m);        for(int i=0; i<n; ++i)cost[i][i]=0;        for(int i=0; i<m; ++i)        {            scanf("%d%d%d",&x,&y,&c);            --x,--y;            cost[y][x]=cost[x][y]=c;        }        int ans=prim();        if(ans==-1){cout<<"Not Unique!\n";continue;}        int flag=0;        for(int i=0; i<n; ++i)        {            for(int j=i+1; j<n; ++j)                if(!in_MST[i][j])                    if(cost[i][j]==Max_e[i][j])                    {                        cout<<"Not Unique!\n";                        flag=1;                        break;                    }            if(flag)break;        }        if(!flag)        cout<<ans<<endl;    }    return 0;}