【次小生成树】环切性质

次小生成树 求法  
求出最小生成树后，建立最优顺序生成树 
枚举每条不在最小生成树上的边，找出左右端点之间的最大边（即 Kruskal 生
成顺序森林上的 lca 或用倍增），求出删去最大边加入此边后的权值（环切性质）。  
或者排序后加并查集，求出每条边的替代边。 
取最小。 

时间复杂度 o(mlogn) 

蒯自hyc学长ppt

poj1679判断最小生成树是否唯一

注意，此题非常坑爹，要判m=0且m的数据范围没给

code

#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <memory>
#include <cstdio>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <iostream>

using namespace std;

const int maxn = 1000 + 5;
const int maxm = 10000 + 5;

typedef int pint[maxn];
typedef int eint[maxm];
typedef unsigned int uint;
typedef long long int64;
typedef unsigned long long uint64;

template <class T> inline T Sqr(T x) { return x * x; }
template <class T> inline T Abs(T x) { return x > 0 ? x : -x; }
template <class T> inline T Min(T a, T b) { return a < b ? a : b; }
template <class T> inline T Max(T a, T b) { return a > b ? a : b; }
template <class T> inline void Swap(T & a, T & b) { T _; _ = a; a = b; b = _; }
template <class T> inline T Ksm(T a, T b, T m) { T _ = 1; for (; b; b >>= 1, a = a * a 

% m) (b & 1) ? _ = _ * a % m : 0; return _ % m; }

struct node 
{ 
   int u, v, w; 
} a[maxm];

int T, n, m, tot, ans, e;
int ufs[maxn], s[maxm], r;
int f[maxn][30 + 5], g[maxn][30 + 5];
pint edge, vp, dep; eint point, next, worth, ve;

void link(int u, int v, int w)
{
   point[++e] = v; next[e] = edge[u]; edge[u] = e; worth[e] = w;
}

bool cmp(node a, node b)
{
   return a.w < b.w;
}

int find(int x)
{
   if (x != ufs[x]) ufs[x] = find(ufs[x]);
   return ufs[x];
}

void init()
{
   memset(f, 0, sizeof(f));
   memset(g, 0, sizeof(g));
   memset(vp, 0, sizeof(vp));
   memset(ve, 0, sizeof(ve));
   memset(next, 0, sizeof(next));
   memset(edge, 0, sizeof(edge));
   memset(worth, 0, sizeof(worth));
   memset(point, 0, sizeof(point));
   for (int i = 1; i <= n; ++i)
      ufs[i] = i;
   for (int i = 1; i <= m; ++i)
      scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w);
   tot = e = r = 0;
}

void kruskal()
{
   int k = 1;
   sort(a + 1, a + m + 1, cmp);
   for (int i = 1, fx, fy; i <= n - 1; ++i)
   {
      while ((fx = find(a[k].u)) == (fy = find(a[k].v))) ++k;
      tot += a[k].w; ve[k] = 1; s[++r] = k++;
      ufs[fx] = fy;
   }
   ans = INT_MAX;
   for (int i = 1; i <= r; ++i)
      link(a[s[i]].u, a[s[i]].v, a[s[i]].w), link(a[s[i]].v, a[s[i]].u, a[s[i]].w);
}

void dfs(int u, int d)
{
   vp[u] = 1, dep[u] = d;
   for (int i = edge[u]; i; i = next[i])
      if (!vp[point[i]])
         f[point[i]][0] = u, g[point[i]][0] = worth[i], dfs(point[i], d + 1);
}

int lca(int u, int v)
{
   if (dep[u] < dep[v]) Swap(u, v);
   int res = 0, d = dep[u] - dep[v];
   for (int i = 0; 1 << i <= d; ++i)
      if ((d >> i) & 1) res = Max(res, g[u][i]), u = f[u][i];
   if (u == v) return res;
   for (int k = 30; k; )
   {
      bool chk = 1;
      for (int i = k; i >= 0; --i)
         if (f[u][i] && f[u][i] != f[v][i])
         {
            res = Max(res, Max(g[u][i], g[v][i]));
            u = f[u][i], v = f[v][i], chk = 0, k = i;
            break;
         }
      if (chk) break;
   }
   return Max(res, Max(f[u][0], f[v][0]));
}

void work()
{
   dfs(a[s[1]].u, 0);
   for (int j = 1; j <= 30; ++j)
      for (int i = 1; i <= n; ++i)
         f[i][j] = f[f[i][j - 1]][j - 1], g[i][j] = Max(g[i][j - 1], g[f[i][j - 1]][j 

- 1]);

   for (int i = 1; i <= m; ++i)
      if (!ve[i])
      {
         int ww, w;
         ww = lca(a[i].u, a[i].v), w = tot - ww + a[i].w;
         if (w < ans && w >= tot) ans = w;
         if (ans == tot) break;
      }

   if (ans == tot) printf("Not Unique!");
   else printf("%d", tot);
}

int main()
{

   for (scanf("%d", &T); T; --T)
   {
      scanf("%d%d", &n, &m);
      if (m == 0)
      {
         printf("0");
         if (T != 1) puts("");
         continue;
      }
      init(), kruskal(), work();
      if (T != 1) puts("");
   }

   return 0;
}