题目描述
It’s universally acknowledged that there’re innumerable trees in the campus of HUST. Thus a professional tree manager is needed. Your task is to write a program to help manage the trees.
Initially, there are n forests and for the i-th forest there is only the i-th tree in it. Given four kinds of operations.
1 u v, merge the forest containing the u-th tree and the forest containing the v-th tree; 2 u, separate the u-th tree from its forest; 3 u, query the size of the forest which contains the u-th tree;
4 u v, query whether the u-th tree and the v-th tree are in the same forest.
输入描述:
The first line contains an integer T, indicating the number of testcases. In each test case: The first line contains two integers N and Q, indicating the number of initial forests and the number of operations. Then Q lines follow, and each line describes an operation. 输出描述:
For each test cases, the first line should be "Case #i:", where i indicate the test case i.
For each query 3, print a integer in a single line.
For each query 4, print "YES" or "NO" in a single line.
示例1
输入
1
10 8
3 1
4 1 2
1 1 2
3 1
4 1 2
2 1
3 1
4 1 2
输出
Case #1:
1
NO
2
YES
1
NO

题意:对trees有四种操作。

1 u v,将包含u-th树的森林和含有v-th树的森林合并在一起;

u,将u-th树与森林分开;

3 u,查询包含u-th树的森林的大小;

4 u v,查询u-th树和v-th树是否在同一林中。

【分析】:给每个点都给他们造一个编号w,删除点,就相当于该点的编号w改变就可以了,也是相当于构造新点。

【出处】:UVA 11987 (白书267)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<string>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<set>
#define LL long long
#define INF 0xffffff
using namespace std;
const double pi = 2 * acos (0.0);
const int maxn = 2e5+10;
int n,m;
int p[maxn];
int s[maxn];
int w[maxn]; void init()
{
for(int i=1;i<=maxn;i++) //这里必须是maxn,其他的好像不行
{
p[i] = i;
w[i] = i;
s[i] = 1;
}
}
int Find(int x)
{
return x==p[x]?x:p[x]=Find(p[x]);
}
void Union(int x,int y)
{
int fx = Find(x);
int fy = Find(y);
if(fx!=fy)
{
p[fx] = fy;
s[fy] += s[fx];
//printf("size=%d\n",s[fy]);
}
} int main()
{
int T;
cin >> T;
int k =1;
while(T--)
{
printf("Case #%d:\n",k++);
init(); cin >> n >> m;
int cnt = n;
while(m--)
{
int op,u,v;
scanf("%d",&op);
if(op==1)
{
scanf("%d%d",&u,&v);
Union(w[u],w[v]);
}
else if(op==2)
{
scanf("%d",&u);
s[Find(w[u])]--;
w[u] = ++cnt;
}
else if(op==3)
{
scanf("%d",&u);
printf("%d\n",s[Find(w[u])]);
}
else{
scanf("%d%d",&u,&v);
if(Find(w[u])==Find(w[v]))
printf("YES\n");
else printf("NO\n");
}
}
}
return 0;
}

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