POJ1679 The Unique MST(Kruskal)(最小生成树的唯一性)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 27141 | Accepted: 9712 |
Description
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
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
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!
【分析】最小生成树的唯一性,思路是先判断每条边是否有重边,有的话eq=1,否则0.然后第一次求出最小生成树,将结果记录下来,
然后依次去掉第一次使用过的且含有重边的边,再求一次最小生成树,若结果与第一次结果一样,则不唯一。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=;
const int M=;
int n,m,cnt;
int parent[N];
bool flag;
struct man {
int u,v,w;
int eq,used,del;
} edg[N];
bool cmp(man g,man h) {
return g.w<h.w;
}
void init() {
for(int i=; i<=; i++) {
parent[i]=i;
}
}
int Find(int x) {
if(parent[x] != x) parent[x] = Find(parent[x]);
return parent[x];
}//查找并返回节点x所属集合的根节点
void Union(int x,int y) {
x = Find(x);
y = Find(y);
if(x == y) return;
parent[y] = x;
}//将两个不同集合的元素进行合并
int Kruskal() {
init();
int sum=;
int num=;
for(int i=;i<m;i++){
if(edg[i].del==)continue;
int u=edg[i].u;int v=edg[i].v;int w=edg[i].w; if(Find(u)!=Find(v)){
sum+=w;
if(!flag)edg[i].used=;
num++;
Union(u,v);
}
if(num>=n-)break;
}
return sum;
}
int main() {
int t,d;
cin>>t;
while(t--) {
cnt=;
cin>>n>>m;
for(int i=; i<m; i++) {
cin>>edg[i].u>>edg[i].v>>edg[i].w;
edg[i].del=;
edg[i].used=;
edg[i].eq=;//一开始这个地方eq没有初始化,WA了好几发,操
}
for(int i=;i<m;i++){
for(int j=;j<m;j++){
if(i==j)continue;
if(edg[i].w==edg[j].w)edg[i].eq=;
}
}
sort(edg,edg+m,cmp);
flag=false;
cnt=Kruskal();
flag=true;
bool gg=false;
for(int i=;i<m;i++){
if(edg[i].used==&&edg[i].eq==){
edg[i].del=;
int s=Kruskal();//printf("%d %d\n",i,s);
if(s==cnt){
gg=true;
printf("Not Unique!\n");
break;
}
edg[i].del=;
}
}
if(!gg)cout<<cnt<<endl;
}
return ;
}
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