POJ 1679 判断最小树是否唯一

来源：互联网发布：2016淘宝怎么开店步骤编辑：程序博客网时间：2024/06/06 11:50

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 1000000000using 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;}

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