算法设计 第二次上机 The Unique MST

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

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'.

输入

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.

输出

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

样例输入

23 31 2 12 3 23 1 34 41 2 22 3 23 4 24 1 2

样例输出

3Not Unique!

来源

POJ Monthly--2004.06.27 srbga@POJ

需要生成最小生成树之后，遍历去掉树上的每一条边，看生成的最小生成树是否与原本的最小生成树权重相同。

注意需要考虑当图是一棵树，并且每条边的长度都为0时的情况。

sort()默认从小到大排序。

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <climits>#include <vector>#include <algorithm>using namespace std;struct node{public:    int u, v, w;    bool operator < (node b)    {        return w < b.w;    }}graph[10005];int father[105];bool MST[10005];int n, m;int findfather(int x){    if(father[x] != x)    {        father[x] = findfather(father[x]);    }    return father[x];}int kruskal(int e){    int cost = 0;    int num = 0;    for(int j = 1; j <= n; j++)    {        father[j] = j;    }    for(int j = 0; j < m; j++)    {        if(num >= n - 1)            break;        if(j == e)            continue;        int x = graph[j].u;        int y = graph[j].v;        if(findfather(x) != findfather(y))        {            father[father[x]] = father[y];            cost += graph[j].w;            num++;            //cout << "j:" << j << "\tcost:" << cost << endl;            if(e == -1)                MST[j] = true;        }    }    return cost;}int main(){    //freopen("test.txt", "r", stdin);    ios::sync_with_stdio(false);    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        memset(MST, false, sizeof(MST));        for(int i = 0; i < m; i++)        {            int xi, yi, wi;            scanf("%d%d%d", &xi, &yi, &wi);            graph[i].u = xi;            graph[i].v = yi;            graph[i].w = wi;        }        sort(graph, graph + m);        int cost1 = kruskal(-1);        int cost2;        bool flag = false;        if(m == n - 1)        {            printf("%d\n", cost1);            continue;        }        for(int i = 0; i < m; i++)        {            if(MST[i])            {                cost2 = kruskal(i);                if(cost1 == cost2)                {                    flag = true;                    break;                }            }        }        if(flag)            printf("Not Unique!\n");        else            printf("%d\n", cost1);    }    return 0;}