poj 1679 The Unique MST （次小生成树）

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!

题意

t(t <= 20)组数据，每组数据给出n，m（n <= 100），表示有n个点，m条边。每条边给出x，y，w表示x到y有一条权值为w的边。问最小生成树是否只有唯一解。

题解

判断最小生成树是否唯一的方法很简单：看次小生成树的值是否与它相等~所以跑一个很朴素的复杂度挺高的次小生成树即可。

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1010;struct Node {    int u, v, w, next;} e[N], b[N];int t, n, m, num, ans, sum, tot, fa[N], head[N], dis[N][N];bool cmp(const Node &a, const Node &b) {    return a.w < b.w;}void add(int u, int v, int w) {    num ++;    e[num].v = v;    e[num].w = w;    e[num].next = head[u];    head[u] = num;}int find(int x) {    if(x == fa[x]) return x;    return fa[x] = find(fa[x]);}void kruskal() {    sort(b + 1, b + m + 1, cmp);    for(int i = 1; i <= m; i ++) {        int fx = find(b[i].u), fy = find(b[i].v);        if(fx == fy) continue;        fa[fy] = fx; sum += b[i].w; b[i].next = true;        add(b[i].u, b[i].v, b[i].w); add(b[i].v, b[i].u, b[i].w);        if(tot ++ == n - 1) return;    }}void dfs(int now, int u, int f, int maxs) {    dis[now][u] = maxs;    for(int i = head[u]; i; i = e[i].next) {        int v = e[i].v;        if (v != f) dfs(now, v, u, max(maxs, e[i].w));    }}int main() {    scanf("%d", &t);    while(t --) {        sum = num = tot = 0;        memset(b, 0, sizeof(b));        memset(head, 0, sizeof(head));        scanf("%d %d", &n, &m);        for(int i = 1; i <= n; i ++) fa[i] = i;        for(int i = 1; i <= m; i ++)            scanf("%d %d %d", &b[i].u, &b[i].v, &b[i].w);        kruskal(); ans = 0x3ffffff;        for(int i = 1; i <= n; i ++) dfs(i, i, 0, 0);        for(int i = 1; i <= m; i ++)            if(! b[i].next) ans = min(ans, sum + b[i].w - dis[b[i].u][b[i].v]);        if(ans != sum) printf("%d\n", sum);        else printf("Not Unique!\n");    }    return 0;}