Lightoj--1009--Back to Underworld(dfs)

Back to Underworld

Time Limit: 4000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

Description

The Vampires and Lykans are fighting each other to death. The war has become so fierce that, none knows who will win. The humans want to know who will survive finally. But humans are afraid of going to the battlefield.

So, they made a plan. They collected the information from the newspapers of Vampires and Lykans. They found the information about all the dual fights. Dual fight means a fight between a Lykan and a Vampire. They know the name of the dual fighters, but don't
know which one of them is a Vampire or a Lykan.

So, the humans listed all the rivals. They want to find the maximum possible number of Vampires or Lykans.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 105), denoting the number of dual fights. Each of the next n lines will contain two different integers u v (1 ≤ u, v ≤ 20000) denoting there was a fight
between u and v. No rival will be reported more than once.

Output

For each case, print the case number and the maximum possible members of any race.

Sample Input

2

2

1 2

2 3

3

1 2

2 3

4 2

Sample Output

Case 1: 2

Case 2: 3

Source

Problem Setter: Jane Alam Jan

题意：有两个帮派打架，判断两个帮派最多的人数有多少
搜索，因为每次给的都是两个帮派打架的编号，并且同一个战斗不会输出两次，我们可以把每个连通块画出来，然后可以发现，相邻的两层一定不是一个帮派的，所以记录下每个连通块奇数层还有偶数层的人数，每次都取最多

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
int a,b;
int n,v[200000+10],vis[200000+10];
vector<int>G[200000+10];
void dfs(int s,int now)
{
for(int i=0;i<G[s].size();i++)
{
if(G[s][i]!=s)
if(vis[G[s][i]]==0)
{
if(now==1)
{
b++;
vis[G[s][i]]=1;
dfs(G[s][i],0);
}
else
{
a++;
vis[G[s][i]]=1;
dfs(G[s][i],1);
}
}
}
}
int main()
{
int t,k=1;
scanf("%d",&t);
while(t--)
{
int mm=0;
int x,y;
memset(v,0,sizeof(v));
memset(vis,0,sizeof(vis));
scanf("%d",&n);
for(int i=0;i<200000;i++)
G[i].clear();
for(int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
if(x>y)
swap(x,y);
G[x].push_back(y);
G[y].push_back(x);
v[x]=v[y]=1;
mm=max(mm,x);
mm=max(mm,y);
}
int sum=0;
for(int i=1;i<=mm;i++)
{
a=b=0;
if(v[i]&&!vis[i])
{
a++;
vis[i]=1;
dfs(i,1);
sum+=max(a,b);
}
}
printf("Case %d: %d\n",k++,sum);
}
return 0;
}