本文主要是介绍POJ 1679 The Unique MST 次小生成树，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目大意就是问是否有多个权值相同的最小生成树
有两种方法， 一种是枚举删边，然后接着构造最小生成树，但是复杂度比较大
另外一种就比较好了 ， 是求次小生成树的方法， 把生成树上任意两点间的最大边在求最小生成树的同时预处理出来，然后n2的枚举任意两点，如果这两点在最小生成树中不是相邻的，就可以删掉两点间的最大边，换上新边，即他俩之间的直接的边，在邻接矩阵中就是他们的距离。
第一种方法 ，我用克鲁斯卡尔做的，很久前的代码了
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 1005
#define INF 100000000
#define eps 1e-7
using namespace std;
struct wwj
{int u, v, w;int equal;int used;int del;
}edge[10005];
int parent[10005], num[10005];
int n, m;
bool first;
bool cmp(wwj x, wwj y)
{return x.w < y.w;
}
void init()
{scanf("%d%d", &n, &m);for(int i = 1; i <= m; i++){scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);edge[i].del = 0; edge[i].equal = 0; edge[i].used = 0;}for(int i = 1; i <= m; i++){for(int j = 1; j <= m; j++){if(i == j) continue;if(edge[j].w == edge[i].w)edge[i].equal = 1;}}sort(edge + 1, edge + m + 1, cmp);first = true;
}
int find(int x)
{if(parent[x] == x)return x;int t = find(parent[x]);parent[x] = t;return t;
}
void join(int x, int y)
{int fx = find(x);int fy = find(y);if(fx != fy){if(num[fx] > num[fy]){parent[fy] = fx;num[fx] += num[fy];num[fy] = 1;}else{parent[fx] = fy;num[fy] += num[fx];num[fx] = 1;}}
}
int kruskal()
{int i, sum = 0, cnt = 0, u, v;for(i = 1; i <= n; i++)parent[i] = i;memset(num, 1, sizeof(num));for(i = 1; i <= m; i++){if(edge[i].del == 1) continue;u = edge[i].u;v = edge[i].v;if(find(u) != find(v)){sum += edge[i].w;cnt++;join(u, v);if(first) edge[i].used = 1;}if(cnt >= n - 1) break;}return sum;
}
void solve()
{int w1 = kruskal(), w2, i;first = false;for(i = 1; i <= m; i++){if(edge[i].used && edge[i].equal){edge[i].del = 1;w2 = kruskal();if(w1 == w2){printf("Not Unique!\n");return;}edge[i].del = 0;}}if(i > m) printf("%d\n", w1);
}
int main()
{
#ifdef LOCALfreopen("d:/data.in","r",stdin);freopen("d:/data.out","w",stdout);
#endifint t;scanf("%d", &t);while(t--){init();solve();}return 0;
}第二种方法是用prim做的
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#define MAXN 1005
#define MAXM 100005
#define INF 1000000000
using namespace std;
int mx[105][105];
int d[105][105];
int can[105][105];
int dis[105], used[105], near[105];
int n, m;
void init()
{for(int i = 1; i <= n; i++){near[i] = 1;for(int j = 1; j <= n; j++){mx[i][j] = 0;d[i][j] = INF;can[i][j] = 0;}d[i][i] = 0;}memset(used, 0, sizeof(used));
}
int main()
{int T, x, y, w;scanf("%d", &T);while(T--){scanf("%d%d", &n, &m);init();for(int i = 1; i <= m; i++){scanf("%d%d%d", &x, &y, &w);d[x][y] = d[y][x] = w;}for(int i = 1; i <= n; i++)dis[i] = d[1][i];used[1] = true;near[1] = -1;int sum = 0;for(int i = 1; i < n; i++){int mi = INF;int v = -1;for(int j = 1; j <= n; j++){if(near[j] != -1 && dis[j] < mi){v = j;mi = dis[j];}}int pre;if(v != -1){pre = near[v];sum += d[v][near[v]];can[v][near[v]] = can[near[v]][v] = 1;mx[near[v]][v] = mx[v][near[v]] = d[v][near[v]];near[v] = -1;for(int j = 1; j <= n; j++){if(near[j] != -1 && d[v][j] < dis[j]){dis[j] = d[v][j];near[j] = v;}}}for(int j = 1; j <= n; j++){if(near[j] == -1 && j != v){mx[j][v] = mx[v][j] = max(mx[j][pre], mx[pre][v]);}}}bool flag = false;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){if(i != j && !can[i][j] && d[i][j] < INF){int sum2 = sum - mx[i][j] + d[i][j];if(sum2 == sum) flag = true;}}if(flag) printf("Not Unique!\n");else printf("%d\n", sum);}return 0;
}
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