POJ 1679 The Unique MST 【用次小生成树验证最小生成树是否唯一】
2016-07-29 15:20
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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
题目大意:给你一棵树,判断他的最小生成树是否为唯一; 方法就是求最小生成树的次小生成树 ,如果加边过程得到的权值和之前的一样,则定然不唯一;
题解:先求出最小生成树,然后不断用未添加进去的最大边去替换去替换树内边, 如果替换过程中产生了一样的权值,则说明最小生成树不唯一;
AC代码(转自kuangbin巨巨,本萌暂时不会这个题,只知道意思)
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!
题目大意:给你一棵树,判断他的最小生成树是否为唯一; 方法就是求最小生成树的次小生成树 ,如果加边过程得到的权值和之前的一样,则定然不唯一;
题解:先求出最小生成树,然后不断用未添加进去的最大边去替换去替换树内边, 如果替换过程中产生了一样的权值,则说明最小生成树不唯一;
AC代码(转自kuangbin巨巨,本萌暂时不会这个题,只知道意思)
//============================================================================ // Name : POJ.cpp // Author : kuangbin // Version : // Copyright : Your copyright notice // Description : Hello World in C++, Ansi-style //============================================================================ #include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; /* * 次小生成树 * 求最小生成树时,用数组Max[i][j]来表示MST中i到j最大边权 * 求完后,直接枚举所有不在MST中的边,替换掉最大边权的边,更新答案 * 点的编号从0开始 */ const int MAXN=110; const int INF=0x3f3f3f3f; bool vis[MAXN]; int lowc[MAXN]; int pre[MAXN]; int Max[MAXN][MAXN];//Max[i][j]表示在最小生成树中从i到j的路径中的最大边权 bool used[MAXN][MAXN]; int Prim(int cost[][MAXN],int n) { int ans=0; memset(vis,false,sizeof(vis)); memset(Max,0,sizeof(Max)); memset(used,false,sizeof(used)); vis[0]=true; pre[0]=-1; for(int i=1;i<n;i++) { lowc[i]=cost[0][i]; pre[i]=0; } lowc[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>lowc[j]) { minc=lowc[j]; p=j; } if(minc==INF)return -1; ans+=minc; vis[p]=true; used[p][pre[p]]=used[pre[p]][p]=true; for(int j=0;j<n;j++) { if(vis[j])Max[j][p]=Max[p][j]=max(Max[j][pre[p]],lowc[p]); if(!vis[j]&&lowc[j]>cost[p][j]) { lowc[j]=cost[p][j]; pre[j]=p; } } } return ans; } int ans; int smst(int cost[][MAXN],int n) { int Min=INF; for(int i=0;i<n;i++) for(int j=i+1;j<n;j++) if(cost[i][j]!=INF && !used[i][j]) { Min=min(Min,ans+cost[i][j]-Max[i][j]); } if(Min==INF)return -1;//不存在 return Min; } int cost[MAXN][MAXN]; int main() { int T; int n,m; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); int u,v,w; for(int i=0;i<n;i++) for(int j=0;j<n;j++) { if(i==j)cost[i][j]=0; else cost[i][j]=INF; } while(m--) { scanf("%d%d%d",&u,&v,&w); u--;v--; cost[u][v]=cost[v][u]=w; } ans=Prim(cost,n); if(ans==-1) { printf("Not Unique!\n"); continue; } if(ans==smst(cost,n))printf("Not Unique!\n"); else printf("%d\n",ans); } return 0; }
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