poj 1679 The Unique MST （次小生成树）

poj 1679 The Unique MST

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!

题目大意：判断一张图的最小生成树的权值和是不是唯一的。

解题思路：找出该图的次小生成树，看最小生成树和次小生成树的权值和是否相同。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;

typedef long long ll;
const int N = 105;
const int INF = 0x3f3f3f3f;
int n, m;
int Max[N][N];
bool vis[N], used[N][N];
int dis[N][N], lowc[N];
int pre[N];

void init() {
    memset(vis, false, sizeof(vis));
    memset(used, false, sizeof(used));
    vis[1] = true;
    pre[1] = -1;
    lowc[1] = 0;
    for (int i = 2; i <= n; i++) {
        lowc[i] = dis[1][i];
        pre[i] = 1;
    }
}   

void input() {
    memset(dis, INF, sizeof(dis));
    int a, b, c;
    for (int i = 0; i < m; i++) {
        scanf("%d %d %d", &a, &b, &c);  
        dis[a][b] = dis[b][a] = c;
    }
}

int prim() {
    int ans = 0;  
    for (int i = 2; i <= n; i++) {  
        int minc = INF;    
        int p = 1;  
        for (int j = 1; j <= n; j++) {  
            if (!vis[j] && minc > lowc[j]) {  
                minc = lowc[j];   
                p = j;  
            }     
        }  
        if (pre[p] != -1) {  
            used[pre[p]][p] = used[p][pre[p]] = true;     
            for (int j = 1; j <= n; j++) {  
                if (vis[j]) {  
                    Max[j][p] = Max[p][j] = max(Max[j][pre[p]], lowc[p]);     
                }     
            }  
        }  
        ans += minc;  
        vis[p] = true;  
        for (int j = 1; j <= n; j++) {  
            if (!vis[j] && lowc[j] > dis[p][j]) {  
                lowc[j] = dis[p][j];  
                pre[j] = p;  
            }     
        }  
    }  
    return ans;  
}

void solve() {
    int MLT = prim();
    int SMLT = INF;
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            if (!used[i][j] && dis[i][j] != INF) {
                SMLT = min(SMLT, MLT + dis[i][j] - Max[i][j]);              
            }   
        }   
    }
    if (MLT == SMLT) printf("Not Unique!\n");
    else printf("%d\n", MLT);
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d %d", &n, &m); 
        input();
        init();
        solve();
    }
    return 0;
}