POJ1679 The Unique MST
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 26782 | Accepted: 9598 |
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!
Source
先跑一遍最小生成树,把树上边都记录下来。
然后枚举不使用其中一条边而再跑最小生成树,若答案没变,说明最小生成树不止一条。
注意数组大小←至少有十道题死在这个问题上了
用n估算的话5000最保险,实际上3000可AC
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int mxn=;
int n,m;
struct edge{
int x,y;
int v;
}e[mxn];
int tot;
int mst[mxn],cnt;
int cmp(const edge a,const edge b){
return a.v<b.v;
}
int fa[mxn];
void init(int x){
for(int i=;i<=x;i++)fa[i]=i;return;
}
int find(int x){
if(fa[x]==x)return x;
return fa[x]=find(fa[x]);
}
void Kruskal(){ init(n);
int i,j;
cnt=;
int ans1=;
tot=;
for(i=;i<=m;i++){
int x=find(e[i].x);int y=find(e[i].y);
if(x!=y){
fa[x]=y;
ans1+=e[i].v;
mst[++cnt]=i;
tot++;
}
if(tot==n-)break;
}
//
int ans2;
for(int k=;k<=cnt;k++){
tot=; ans2=;
init(n);
//init
for(i=;i<=m;i++){
if(i==mst[k])continue;
int x=find(e[i].x);int y=find(e[i].y);
if(x!=y){
fa[x]=y;
ans2+=e[i].v;
// mst[++cnt]=i;//!!这步不能加! 偷懒从上面复制的结果就是WA记录喜+1
tot++;
}
if((tot==n-) && ans1==ans2){
printf("Not Unique!\n");
return;
}
}
}
printf("%d\n",ans1);
return;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++)
scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].v);
sort(e+,e+m+,cmp);
Kruscal();
}
return ;
}
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