题意: 给你一个二分图, (可能不连通) 求可能多的子集元素个数;
思路: 直接DFS 给二分图染色就有了, 统计联通块中个数, 去最大值相加即可。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 20000 + 131;
struct Node {
int One, Zre;
}N[maxn / 2];
vector<int> E[maxn];
bool Vis[maxn];
int Cnt ;
void DFS(int e, int k)
{
for(int i = 0; i < E[e].size(); ++i)
{
if(Vis[E[e][i]] == false)
{
//cout << "This is E : " << E[e][i] << endl;
Vis[E[e][i]] = true;
if(k == 0) N[Cnt].Zre ++;
else N[Cnt].One ++;
DFS(E[e][i], !k);
}
}
}
int Solve()
{
Cnt = 0;
for(int i = 0; i < maxn; ++i)
{
if(Vis[i] == false && E[i].size())
{
Cnt ++;
N[Cnt].One = 1;
Vis[i] = true;
DFS(i, 0);
}
}
/*for(int i = 1; i <= Cnt; ++i)
{
cout << N[i].Zre << " " << N[i].One << endl;
}*/
int Ans = 0;
for(int i = 1; i <= Cnt; ++i)
Ans += max(N[i].Zre, N[i].One);
return Ans;
}
void INIT()
{
memset(N,0,sizeof(N));
memset(Vis,0,sizeof(Vis));
for(int i = 0; i < maxn; ++i)
E[i].clear();
}
int main()
{
int t, n, u ,v;
scanf("%d", &t);
for(int kase = 1; kase <= t; ++ kase)
{
INIT();
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
scanf("%d %d", &u, &v);
E[u].push_back(v);
E[v].push_back(u);
}
int ret = Solve();
printf("Case %d: %d\n", kase, ret);
}
}
原文:http://www.cnblogs.com/aoxuets/p/4915255.html