【poj 1679】The Unique MST 【次小生成树 模板】

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!

模板连接

题意 ： 给定n个点m条边的图，求最小生成树，如果最小生成树不唯一那么就输出Not Unique!，否则就输出最小生成树的值；
代码

#include<cstdio>#include<string.h>#include<iostream>#include<math.h> using namespace std;#define LL long long const int inf = 0x3f3f3f3f;const int MAXN =200+10;const int MAXM =1e8;/*------------------------------*/int mp[MAXN][MAXN];bool vis[MAXN];int dis[MAXN];int mst[MAXN][MAXN];//记录生成树中，i到j路中最大的权值是多少。bool Inmst[MAXN][MAXN];//记录边是不是在生成树中int pre[MAXN];//记录每个点的前驱int n,m;void init(){    for(int i=1;i<=n;i++){        for(int j=1;j<i;j++){            if(i==j ) mp[i][j]=0;            else mp[i][j]=mp[j][i]=inf;        }    }}void getmap(){    int a,b,c;    while(m--){        scanf("%d%d%d",&a,&b,&c);        if(mp[a][b]>c){ //防止重边            mp[a][b]=mp[b][a]=c;        }    }}void solve(){ // prime算法求最小生成树    memset(mst,0,sizeof(mst));    memset(Inmst,0,sizeof(Inmst));    for(int i=1;i<=n;i++){        dis[i]=mp[1][i]; //默认选择 1为起始点        vis[i]=0;        pre[i]=1;    }    vis[1]=1;int minn,nexts;int mincost=0;    for(int i=2;i<=n;i++){          minn=inf;nexts=-1;          for(int j=1;j<=n;j++){            if(!vis[j]&&minn>dis[j]) {                minn=dis[j];                nexts=j;              }          }          vis[nexts]=1;mincost+=minn;int fa=pre[nexts];          Inmst[fa][nexts]=Inmst[nexts][fa]=1;          for(int j=1;j<=n;j++){              if(vis[j]&&j!=nexts) mst[nexts][j]=mst[j][nexts]=max(mp[fa][nexts],dis[j]); //更新 mst              if(!vis[j]&&dis[j]>mp[nexts][j]) {                    dis[j]=mp[nexts][j];                    pre[j]=nexts;//更新前驱                }          }    }    for(int i=1;i<=n;i++){ // 枚举边        for(int j=1;j<i;j++){            if(mp[i][j]!=inf&&!Inmst[i][j]){//有这个边，但是却不在生成树中                if(mp[i][j]==mst[i][j]){//可以被替代                    puts("Not Unique!");                    return ;                }             }        }    }    printf("%d\n",mincost);}int main(){        int t;cin>>t;        while(t--){             scanf("%d%d",&n,&m);            init();            getmap();            solve();        }    return 0;}