POJ 1679 判断最小树是否唯一

题意：

      给你一个图，问你最小树是否唯一，唯一则输出最小数的权值，不唯一输出Not Unique!

思路：

     题目问的是最小树是否唯一，其实也就是在问次小树是否等于最小树，如果等于则不唯一，求次小树快的方法应该是先求最小树，然后枚举删除最小树上的边，最快的应该是树形dp优化的那个吧，刚刚忘记了，直接求出最小树，然后暴力深搜分成两个集合枚举，0ms AC，因为点才100，所以暴力也无压力。

#include<stdio.h>
#include<string.h>
#include<algorithm>

#define N_node 100 + 10
#define N_edge 20000 + 100
#define INF 1000000000

using namespace std;

typedef struct
{
   int to ,next ,cost;
}STAR;

typedef struct
{
   int a ,b ,c;
}EDGE;

STAR E[N_edge];
EDGE edge[N_edge] ,Tree[N_node];
int list[N_node] ,tot;
int mer[N_node];
int mark[N_node];
int map[N_node][N_node];
int node_l[N_node] ,node_r[N_node];
int ll ,rr;

void add(int a ,int b)
{
    E[++tot].to = b;
    E[tot].next = list[a];
    list[a] = tot;
    E[++tot].to = a;
    E[tot].next = list[b];
    list[b] = tot;
}

bool camp(EDGE a ,EDGE b)
{
     return a.c < b.c;
}

int finds(int x)
{
    return x == mer[x] ? x : mer[x] = finds(mer[x]);
}

void DFS1(int s)
{
   node_l[++ll] = s;
   for(int k = list[s] ;k ;k = E[k].next)
   {
      int to = E[k].to;
      if(mark[to]) continue;
      mark[to] = 1;
      DFS1(to);
   }
}

void DFS2(int s)
{
   node_r[++rr] = s;
   for(int k = list[s] ;k ;k = E[k].next)
   {
      int to = E[k].to;
      if(mark[to]) continue;
      mark[to] = 1;
      DFS2(to);
   }
}

int minn(int x ,int y)
{
     return x < y ? x : y;
} 

int main ()
{
   int t ,n ,m ,i ,j;
   scanf("%d" ,&t);
   while(t--)
   {
      scanf("%d %d" ,&n ,&m);
      for(i = 1 ;i <= n ;i ++)
      for(j = 1 ;j <= n ;j ++)
      map[i][j] = INF;
      for(i = 1 ;i <= m ;i ++)
      {
         scanf("%d %d %d" ,&edge[i].a ,&edge[i].b ,&edge[i].c);
         map[edge[i].a][edge[i].b] = map[edge[i].b][edge[i].a] = edge[i].c;
      }
      sort(edge + 1 ,edge + m + 1 ,camp);
      int sum = 0;
      for(i = 1 ;i <= n ;i ++) mer[i] = i;
      memset(list ,0 ,sizeof(list)) ,tot = 1;
      int tt = 0;
      for(i = 1 ;i <= m ;i ++)
      {
         int x = finds(edge[i].a);
         int y = finds(edge[i].b);
         if(x == y) continue;
         mer[x] = y;
         sum += edge[i].c;
         add(edge[i].a ,edge[i].b);
         Tree[++tt].a = edge[i].a;
         Tree[tt].b = edge[i].b;
         Tree[tt].c=  edge[i].c;
      }
      int now = 1000000000;
      for(i = 1 ;i <= tt ;i ++)
      {
         int a = Tree[i].a;
         int b = Tree[i].b;
         int c = Tree[i].c;
         memset(mark ,0 ,sizeof(mark));
         mark[a] = mark[b] = 1;
         ll = 0;
         DFS1(a);
         rr = 0;
         DFS2(b);
         for(int ii = 1 ;ii <= ll ;ii ++)
         for(int jj = 1 ;jj <= rr ;jj ++)
         {
            int aa = node_l[ii];
            int bb = node_r[jj];
            if(map[aa][bb] == INF || aa == a && bb == b || aa == b && bb == a) continue;
            now = minn(now ,sum - c + map[aa][bb]);
         }
      }
      if(now != sum) printf("%d\n" ,sum);
      else printf("Not Unique!\n");
   }
   return 0;
}