POJ1679 The Unique MST（次小生成树）

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34321		Accepted: 12513

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:

V' = V.
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

Sample Output

3
Not Unique!

Source

POJ Monthly--2004.06.27 srbga@POJ

链接

http://poj.org/problem?id=1679

题意

给定一个无向带权图，问最小生成树是否唯一。

题解

先用Prim求得其最小生成树ans1，然后求得次小生成树ans2，如果ans1=ans2，那么最小生成树不唯一。

代码

Status	Accepted
Memory	240kB
Length	2130

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 110;
int G[maxn][maxn];
int Max[maxn][maxn];
bool used[maxn][maxn];
bool vis[maxn];
int pre[maxn];
int lowc[maxn];
int n, m;

int Prim() // 求得最小生成树
{
    int ans = 0;
    memset(vis, 0, sizeof(vis));
    memset(Max, 0, sizeof(Max));
    memset(used, 0, sizeof(used));
    vis[1] = true;
    pre[1] = -1;
    lowc[1] = 0;
    for (int i = 2; i <= n; ++i)
    {
        lowc[i] = G[1][i];
        pre[i] = 1;
    }
    for (int i = 2; i <= n; ++i)
    {
        int minc = inf, p = -1;
        for (int j = 1; j <= n; ++j)
        {
            if (!vis[j] && minc > lowc[j])
            {
                minc = lowc[j];
                p = j;
            }
        }
        if (minc == inf) // 不存在最小生成树
            return -1;
        ans += minc;
        vis[p] = true;
        used[p][pre[p]] = used[pre[p]][p] = true;
        for (int j = 1; j <= n; ++j)
        {
            if (vis[j] && j != p)
                Max[j][p] = Max[p][j] = max(Max[j][pre[p]], lowc[p]); // 最小瓶颈路
            if (!vis[j] && lowc[j] > G[p][j])
            {
                lowc[j] = G[p][j];
                pre[j] = p;
            }
        }
    }
    return ans;
}

int smst(int tmp)
{
    int ans = inf;
    for (int i = 1; i <= n; ++i) // 遍历不在最小生成树中的边
    {
        for (int j = 1; j <= n; ++j)
        {
            if (!used[i][j] && i != j)
                ans = min(ans, tmp + G[i][j] - Max[i][j]);
        }
    }
    if (ans == inf)
        return -1;
    return ans;
}

int main()
{
    int T, u, v, w;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                G[i][j] = ((i == j) ? 0 : inf);
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d%d", &u, &v, &w);
            G[u][v] = G[v][u] = w;
        }
        int ans1 = Prim(), ans2 = smst(ans1);
        if (ans1 == ans2 || ans1 == -1) // 判断是否相等
            printf("Not Unique!\n");
        else
            printf("%d\n", ans1);
    }
    return 0;
}

次小生成树裸题，关键是怎么根据最小生成树求次小生成树，思想转化很重要，同时学习了最小瓶颈路。

The end.
2018-05-11 星期五