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

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! 这题直接暴力枚举,找到一颗最小生成树,标记一下,依次删边
 #include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <string>
#include <math.h>
using namespace std;
const int maxn = ;
const int INF = 0x7fffffff;
struct node {
int u, v, w;
int used, num, flag;
} qu[ * maxn];
int fa[maxn], n, m;
int cmp(node a, node b) {
return a.w < b.w;
}
void init() {
for (int i = ; i <= n ; i++) fa[i] = i;
}
int Find(int x) {
return fa[x] == x ? x : fa[x] = Find(fa[x]);
}
int combine(int x, int y) {
int nx = Find(x);
int ny = Find(y);
if (nx != ny) {
fa[nx] = ny;
return ;
}
return ;
}
int kruskal(int flag) {
init();
int sum = , cnt = ;
for (int i = ; i < m ; i++) {
if (qu[i].flag) continue;
if (combine(qu[i].v, qu[i].u)) {
if (!flag) qu[i].used = ;
sum += qu[i].w;
cnt++;
if (cnt == n - ) break;
}
}
if (cnt != n - ) return -;
return sum;
}
int main() {
int t;
scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
for (int i = ; i < m ; i++) {
scanf("%d%d%d", &qu[i].u, &qu[i].v, &qu[i].w);
qu[i].flag = , qu[i].used = ;
}
sort(qu, qu + m, cmp);
int sum = kruskal();
int flag = ;
for (int i = ; i < m ; i++) {
if (qu[i].used ) {
qu[i].flag = ;
int temp = kruskal();
qu[i].flag = ;
if (temp == sum) {
flag = ;
break;
}
}
}
if (flag) printf("Not Unique!\n");
else printf("%d\n", sum);
}
return ;
}

poj1679 次最小生成树 kruskal(暴力枚举)的更多相关文章

  1. poj1679(最小生成树)

    传送门:The Unique MST 题意:判断最小生成树是否唯一. 分析:先求出原图的最小生成树,然后枚举删掉最小生成树的边,重做kruskal,看新的值和原值是否一样,一样的话最小生成树不唯一. ...

  2. CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution (暴力枚举)

    题意:求定 n 个数,求有多少对数满足,ai^bi = x. 析:暴力枚举就行,n的复杂度. 代码如下: #pragma comment(linker, "/STACK:1024000000 ...

  3. 2014牡丹江网络赛ZOJPretty Poem(暴力枚举)

    /* 将给定的一个字符串分解成ABABA 或者 ABABCAB的形式! 思路:暴力枚举A, B, C串! */ 1 #include<iostream> #include<cstri ...

  4. HNU 12886 Cracking the Safe(暴力枚举)

    题目链接:http://acm.hnu.cn/online/?action=problem&type=show&id=12886&courseid=274 解题报告:输入4个数 ...

  5. 51nod 1116 K进制下的大数 (暴力枚举)

    题目链接 题意:中文题. 题解:暴力枚举. #include <iostream> #include <cstring> using namespace std; ; ; ch ...

  6. Codeforces Round #349 (Div. 1) B. World Tour 最短路+暴力枚举

    题目链接: http://www.codeforces.com/contest/666/problem/B 题意: 给你n个城市,m条单向边,求通过最短路径访问四个不同的点能获得的最大距离,答案输出一 ...

  7. bzoj 1028 暴力枚举判断

    昨天梦到这道题了,所以一定要A掉(其实梦到了3道,有两道记不清了) 暴力枚举等的是哪张牌,将是哪张牌,然后贪心的判断就行了. 对于一个状态判断是否为胡牌,1-n扫一遍,然后对于每个牌,先mod 3, ...

  8. POJ-3187 Backward Digit Sums (暴力枚举)

    http://poj.org/problem?id=3187 给定一个个数n和sum,让你求原始序列,如果有多个输出字典序最小的. 暴力枚举题,枚举生成的每一个全排列,符合即退出. dfs版: #in ...

  9. hihoCoder #1179 : 永恒游戏 (暴力枚举)

    题意: 给出一个有n个点的无向图,每个点上有石头数个,现在的游戏规则是,设置某个点A的度数为d,如果A点的石子数大于等于d,则可以从A点给每个邻接点发一个石子.如果游戏可以玩10万次以上,输出INF, ...

随机推荐

  1. linux特殊权限位suid

    特殊权限位基本说明(了解): linux系统基本权限位为9位权限,但还有额外3位权限位,共12位权限: suid       s(x)     S     4     用户对应的权限位(用户对应的3位 ...

  2. vue本人常用插件汇总(常更新)

    1. 移动端UI插件 mint-ui http://mint-ui.github.io/#!/zh-cn 2.vue状态管理vuex,持久化插件:vuex-persist https://github ...

  3. Charles Dickens【查尔斯·狄更斯】

    Charles Dickens In 1812, the year Charles Dickens was born, there were 66 novels published in Britai ...

  4. POJ 3581 三段字符串(后缀数组)

    Sequence Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7923   Accepted: 1801 Case Tim ...

  5. POJ 2084 Catalan

    Game of Connections Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8772   Accepted: 43 ...

  6. Codeforces Round #449 (Div. 2) C. DFS

    C. Nephren gives a riddle time limit per test 2 seconds memory limit per test 256 megabytes input st ...

  7. Pandas库入门

    pandas库的series类型

  8. PHP.21-商品信息管理

    商品信息管理 在线增删改查和图片信息管理 主要技术:文件上传.图片缩放.数据库基本操作 思路: 1.设计并创建数据库 库名:demodb 表名:goods 编号(id) 名称(name) 商品类型(t ...

  9. [转载]python 变量命名规范

    原文地址:python 变量命名规范作者:loveflying python源码和其他一些书籍,命名各种个性,没有一个比较统一的命名规范.于是自己总结了一些,可供参考. 模块名: 小写字母,单词之间用 ...

  10. 过滤器(Filter)和Nuget

    一.过滤器 AOP(面向切面编程)是一种架构思想,用于把公共的逻辑放到一个单独的地方,这样就不用每个地方都写重复的代码,比如程序中发生异常,不用每个地方都try catch 只要在(golbal的Ap ...