1198 C Matching vs Independent Set
大意: 给定$3n$个点的无向图, 求构造$n$条边的匹配, 或$n$个点的独立集.
假设已经构造出$x$条边的匹配, 那么剩余$3n-2x$个点, 若$x<n$, 则$3n-2x\ge n$可以构造出独立集.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int u[N], v[N], a[N], vis[N]; void work() { int n, m; scanf("%d%d", &n, &m); int cnt = 0; REP(i,1,3*n) vis[i] = 0; REP(i,1,m) { scanf("%d%d", u+i, v+i); if (!vis[u[i]]&&!vis[v[i]]&&cnt<n) { a[i]=vis[u[i]]=vis[v[i]]=1; ++cnt; } else a[i] = 0; } if (cnt==n) { puts("Matching"); REP(i,1,m) if (a[i]) printf("%d ",i); } else { puts("IndSet"); cnt = 0; REP(i,1,3*n) if (!vis[i]) { printf("%d ",i); if (++cnt==n) break; } } puts(""); } int main() { int t; scanf("%d", &t); while (t--) work(); }