Connectivity

6492: Connectivity

时间限制: 1 Sec  内存限制: 128 MB
提交: 118  解决: 28
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题目描述

There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the pi-th and qi-th cities, and the i-th railway bidirectionally connects the ri-th and si-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.
We will say city A and B are connected by roads if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define connectivity by railways similarly.
For each city, find the number of the cities connected to that city by both roads and railways.

Constraints
2≤N≤2*105
1≤K,L≤105
1≤pi,qi,ri,si≤N
pi<qi
ri<si
When i≠j, (pi,qi)≠(pj,qj)
When i≠j, (ri,si)≠(rj,sj)

输入

The input is given from Standard Input in the following format:
N K L
p1 q1
:
pK qK
r1 s1
:
rL sL

输出

Print N integers. The i-th of them should represent the number of the cities connected to the i-th city by both roads and railways.

样例输入

4 3 1
1 2
2 3
3 4
2 3

样例输出

1 2 2 1
统计两个并查集交集所含元素的个数，这个数就是交集中的每一个元素对应的答案（实际上是由每一个元素得到总的个数）
还有就是熟练掌握map和pair的用法，我就是知道思路但不知道怎么实现！
AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=2e5+100;
#define p make_pair
map<pair<int,int>,int>mapp;
int sa[maxn],sb[maxn];
struct record
{
int fa[maxn];
void unit(int n)
{
for(int i=1;i<=n;i++)
{
fa[i]=i;
}
}
int finding(int a)
{
if(fa[a]==a)
{
return a;
}
return fa[a]=finding(fa[a]);
}
void union_(int x,int y)
{
int b1=finding(x);
int b2=finding(y);
if(b1!=b2)
{
fa[b1]=b2;
}
}
};
int main()
{
int n,k,l,x,y;
scanf("%d %d %d",&n,&k,&l);
record s1,s2;
s1.unit(n);
s2.unit(n);
for(int i=1;i<=k;i++)
{
scanf("%d %d",&x,&y);
s1.union_(x,y);
}
for(int i=1;i<=l;i++)
{
scanf("%d %d",&x,&y);
s2.union_(x,y);
}
for(int i=1; i<=n; i++)
{
sa[i]=s1.finding(i);
sb[i]=s2.finding(i);
mapp[p(sa[i],sb[i])]++;
}
for(int i=1;i<=n;i++)
{
if(i==n)
{
printf("%d\n",mapp[p(sa[i],sb[i])]);
}
else
{
printf("%d ",mapp[p(sa[i],sb[i])]);
}
}
return 0;
}