lightoj 1009 - Back to Underworld（二分图染色）

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

Output for Sample Input

2

2

1 2

2 3

3

1 2

2 3

4 2

Case 1: 2

Case 2: 3

题意是给你n组关系，代表这两个数是对立关系，然后让你判断最多的一组里最多有可能有多少个人。

思路就是用二分图染色，然而这题的坑点是下标不是连续的，图可能不是联通的，所以做了一些处理。

#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn = 20005;int a,b;int v[200005];int color[maxn];vector<int> e[maxn];void dfs(int x,int c){    color[x] = c;    if(c == 1)        a++;    if(c == -1)        b++;    for(int i=0;i<e[x].size();i++)    {        int t = e[x][i];        if(!color[t])            dfs(t,-c);    }    return;}int main(void){    int T,n,i,j;    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%d",&n);        memset(color,0,sizeof(color));        for(i=0;i<=maxn;i++)            e[i].clear();        int cnt = 0;        for(i=1;i<=n;i++)        {            int x,y;            scanf("%d%d",&x,&y);            e[x].push_back(y);            e[y].push_back(x);            v[cnt++] = x;            v[cnt++] = y;        }        sort(v,v+cnt);        cnt = unique(v,v+cnt) - v;        int ans = 0;        for(i=0;i<cnt;i++)            if(!color[v[i]])            {                a = b = 0;                dfs(v[i],1);                ans += max(a,b);            }        printf("Case %d: %d\n",cas++,ans);    }    return 0;}