poj-1679 The Unique MST（次小生成树模板）

The Unique MST

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 26460 Accepted: 9462

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!

题意：图中最小生成树是否唯一。唯一输出最小生成树长度，不唯一输出Not Unique!

思路：求出次小生成树，然后比较下是否相等，相等则不唯一。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 110#define INF 1<<30int ma[N][N];///ma数组用来保存初始图int p[N],d[N],path[N][N];///p是每个城市的人口，d用来维护当前最短距离,path用来记录最小生成树上两点之间所有路径中的单边最大权值int flag[N],pre[N],n,used[N][N];///flag数组标记当前点是否已经在最小生成树里,pre记录当前点的前驱（因为是无向图，其实也就是一条边的另一个节点）///used[i][j]用来记录边ij有没有在最小生成树上,0是不在，1是在int prim(){    int ret=0;///最小生成树大小    memset(flag,0,sizeof(flag));    memset(used,0,sizeof(used));    memset(path,0,sizeof(path));///题目边都是大于0的，所以置为0就好    for(int i=1; i<n; i++)    {        d[i]=ma[0][i];        pre[i]=0;    }    flag[0]=1;    for(int i=1; i<n; i++)    {        int v=-1;        for(int j=0; j<n; j++)            if(!flag[j]&&(v==-1||(d[v]>d[j])))                v=j;        if(v==-1)            return -1;        used[v][pre[v]]=used[pre[v]][v]=1;        ret+=d[v];        flag[v]=1;        for(int j=0; j<n; j++)        {            if(flag[j]&&j!=v)                path[j][v]=path[v][j]=max(path[pre[v]][j],d[v]);            if(!flag[j]&&d[j]>ma[v][j])            {                d[j]=ma[v][j];                pre[j]=v;            }        }    }    return ret;}int main(){    int T,m;    int u,v,w;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&m);        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                ma[i][j]=INF;        for(int i=0; i<m; i++)        {            scanf("%d %d %d",&u,&v,&w);            ma[u-1][v-1]=ma[v-1][u-1]=w;        }        int ret=prim();        int ans=INF;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                if(i==j) continue;                if(ma[i][j]!=INF&&!used[i][j])                        ans=min(ans,ret-path[i][j]+ma[i][j]);            }        }        if(ans==ret)             printf("Not Unique!\n");        else            printf("%d\n",ret);    }    return 0;}

次小生成树模板：

int ma[N][N];///ma数组用来保存初始图int p[N],d[N],path[N][N];///p是每个城市的人口，d用来维护当前最短距离,path用来记录最小生成树上两点之间所有路径中的单边最大权值int flag[N],pre[N],n,used[N][N];///flag数组标记当前点是否已经在最小生成树里,pre记录当前点的前驱（因为是无向图，其实也就是一条边的另一个节点）///used[i][j]用来记录边ij有没有在最小生成树上,0是不在，1是在int prim(){    int ret=0;///最小生成树大小    memset(flag,0,sizeof(flag));    memset(used,0,sizeof(used));    memset(path,0,sizeof(path));///题目边都是大于0的，所以置为0就好    for(int i=1; i<n; i++)    {        d[i]=ma[0][i];        pre[i]=0;    }    flag[0]=1;    for(int i=1; i<n; i++)    {        int v=-1;        for(int j=0; j<n; j++)            if(!flag[j]&&(v==-1||(d[v]>d[j])))                v=j;        if(v==-1)            return -1;        used[v][pre[v]]=used[pre[v]][v]=1;///标记最小生成树上的边        ret+=d[v];        flag[v]=1;        for(int j=0; j<n; j++)        {            if(flag[j]&&j!=v)                path[j][v]=path[v][j]=max(path[pre[v]][j],d[v]);            if(!flag[j]&&d[j]>ma[v][j])            {                d[j]=ma[v][j];                pre[j]=v;            }        }    }    return ret;}int next(int ret)///求次小生成树长度{    int ans=INF;    for(int i=0; i<n; i++)    {        for(int j=0; j<n; j++)        {            if(i==j) continue;            if(ma[i][j]!=INF&&!used[i][j])                ans=min(ans,ret-path[i][j]+ma[i][j]);        }    }    return ans;}