POJ 1679 The Unique MST (次小生成树)

题目链接：POJ 1679

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!

Source

POJ Monthly—2004.06.27 srbga@POJ

Solution

题意

给定一个连通的无向图 (没有重边)，求最小生成树是否唯一。

思路

Kruskal

求出次小生成树，如果次小生成树的边权和与最小生成树的边权和相等，则最小生成树不唯一。

次小生成树的求法有很多，这里给出其中一种。

次小生成树与最小生成树仅有一条边不同。先用 Kruskal 求出最小生成树，然后枚举最小生成树的每一条边，求删去该边后的最小生成树，所有情况中的最小的生成树就是次小生成树。

Code

#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 110, M = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int n, m;

struct Edge {
    int x, y, z;
} edge[M];

int fa[N];

int cmp(Edge a, Edge b) {
    return a.z < b.z;
}

int get(int x) {
    if(x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}

void init() {
    for(int i = 0; i <= N; ++i) {
        fa[i] = i;
    }
}

int mst[M];  // 最小生成树所有边的下标
int mst_size;  // 最小生成树的边权和

int kruskal() {
    init();
    int flag = 0;
    int ans = 0;
    mst_size = 0;
    sort(edge + 1, edge + 1 + m, cmp);
    for(int i = 1; i <= m; ++i) {
        int x = get(edge[i].x);
        int y = get(edge[i].y);
        if(x != y) {
            ans += edge[i].z;
            fa[x] = y;
            mst[++mst_size] = i;
        }
    }
    return ans;
}

// 次小生成树
int sec_mst() {
    int ans = inf;
    for(int i = 1; i <= mst_size; ++i) {
        init();
        int sum = 0;
        int cnt = 0;
        for(int j = 1; j <= m; ++j) {
            if(j != mst[i]) {  // 删去最小生成树中的一条边
                int x = get(edge[j].x);
                int y = get(edge[j].y);
                if(x != y) {
                    sum += edge[j].z;
                    fa[x] = y;
                    ++cnt;
                }
            }
        }
        if(cnt != mst_size) continue;
        ans = min(ans, sum);
    }
    if(ans == inf) return -1;
    return ans;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= m; ++i) {
            scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].z);
        }
        int tmp1 = kruskal();
        int tmp2 = sec_mst();
        if(tmp1 == tmp2) {
            printf("Not Unique!\n");
        } else {
            printf("%d\n", tmp1);
        }
    }
    return 0;
}