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
" ,sum);
else printf("Not Unique!
");
}
return 0;
}