[2016-01-27][POJ][1679][The Unique MST]
[2016-01-27][POJ][1679][
The Unique MST]
C - The Unique MST
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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'.
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!
- 时间:2016-01-21 11:06:30 星期四
- 题目编号:POJ 1679
- 题目大意:给出一个图,判断它的最小生成树 是否是唯一的.
- 方法:在依次删除一条最小生成树的一条边,重新跑最小生成树,
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#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using
namespace
std;
const
int
maxn = 110;
int
n,m;
struct
Edge{
int
u,v;
long
long
c;
bool
operator < (
const
Edge & a)
const
{
return
c < a.c;
}
}e[maxn*maxn/2];
int
fa[maxn];
void
ini(){
for
(
int
i = 0;i < n + 5;i++)
fa[i] = i;
}
int
fnd(
int
x){
return
x == fa[x] ? x : (fa[x] = fnd(fa[x]));
}
vector<Edge> chs;
int
kruskal()
{
sort(e,e + m);
int
ct = 0;
long
long
res = 0;
ini();
chs.clear();
for
(
int
i = 0;i < m;i++)
{
if
(ct == n - 1)
break
;
int
u = e[i].u,v = e[i].v;
u = fnd(u);
v = fnd(v);
if
(u != v)
{
fa[u] = v;
res += e[i].c;
chs.push_back(e[i]);
ct++;
}
}
/****************************************/
for
(
int
j = 0;j < chs.size();j++)
{
long
long
rres = 0;
ini();
ct = 0;
for
(
int
i = 0;i < m;i++)
{
if
(ct == n - 1)
break
;
if
( chs[j].v == e[i].v && chs[j].u == e[i].u && chs[j].c == e[i].c)
continue
;
int
u = e[i].u,v = e[i].v;
u = fnd(u);
v = fnd(v);
if
(u != v)
{
fa[u] = v;
rres += e[i].c;
ct++;
}
}
if
(rres == res && ct == n - 1)
{
res = -1;
break
;
}
}
/*******************************************/
return
res;
}
int
main(){
int
t;
scanf
(
"%d"
,&t);
while
(t--){
scanf
(
"%d%d"
,&n,&m);
int
a,b,c;
for
(
int
i = 0;i < m ;i++){
scanf
(
"%d%d%d"
,&a,&b,&c);
e[i].u = a;
e[i].v = b;
e[i].c = c;
}
int
res = kruskal();
if
(res != -1)
printf
(
"%d\n"
,res);
else
puts
(
"Not Unique!"
);
}
return
0;
}
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