大家好，又见面了，我是全栈君。
The Unique MST
Time Limit: 1000MS | Memory Limit: 10000K | |
|---|---|---|
Total Submissions: 19941 | Accepted: 6999 |
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
代码语言：javascript复制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 2Sample Output
代码语言：javascript复制3
Not Unique!这道题就是要你推断是否为唯一的最小生成树。。
这也是我第一道次小生成树的题。。那个PDF资料真是太好了。。我用的是Kruskal。。
代码语言：javascript复制#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;
const int max1 = 105;
const int max2 = 10005;
struct node
{
int u, v, w;//边的顶点以及权值
int r;//用于记录边的序号
int used;//用于记录边是否被使用过
}edge[max2];//边的数组
int parent[max1];//顶点i所在集合相应树中的根节点
int n, m;//顶点数，边的个数
int cas;
void start()//初始化
{
for(int i=1; i<=n; i )
parent[i] = i;
}
int find (int x)//查找并返回节点x所属集合的根节点
{
return parent[x] == x ? x : parent[x] = find ( parent[x] );}bool cmp ( node a, node b )//按权值从小到大排序{ return a.w < b.w;}int Kruskal()//第一次求最小生成树{ sort( edge 1, edge m 1, cmp ); int ans = 0; for(int i=1; i<=m; i ) { int e = edge[i].r; int x = find( edge[i].u ); int y = find( edge[i].v ); if( x!=y ) { parent[y] = x; ans = edge[i].w; edge[e].used = 1; } } return ans;}int Kruskal_again(int tt)//求次小生成树{ sort( edge 1, edge m 1, cmp ); int ans = 0; for(int i=1; i<=m; i ) { if( tt==edge[i].r ) continue; int x = find( edge[i].u ); int y = find( edge[i].v ); if( x!=y ) { parent[y] = x; ans = edge[i].w; } } return ans;}bool judge()//推断是否能就得最小生成树，即看是否有孤立的点{ for(int i=1; i<=n; i ) if( find(i) != find(1) ) return false; return true;}int main(){ scanf("%d", &cas); while( cas-- ) { scanf("%d%d", &n, &m); start(); for(int i=1; i<=m; i ) { scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w); edge[i].used = 0; edge[i].r = i; } int flag = 0; int ans1 = Kruskal(); for(int i=1; i<=m; i )//一一枚举求最小生成树。。 { if( !edge[i].used ) continue; start(); int ans2 = Kruskal_again(i);//求除去此边的最小生成树 if( ans1 == ans2 && judge() ) { flag = 1;//标记结论 break; } } if( flag ) printf("Not Unique!n"); else printf("%dn", ans1); } return 0;}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/115989.html原文链接：https://javaforall.cn
