题意:中文题
思路:和poj食物链的题几乎一样,拆点或者带权并查集做,这种分类不多的比较倾向与拆点做
AC代码:
#include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #define ll long long #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a) memset(a,0,sizeof(a)) #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) using namespace std; const long long INF = 1e18+1LL; const int inf = 1e9+1e8; const int N=1e5+100; const ll mod=1e9+7; int pre[N<<1],n,m; void init(int n){ for(int i=0; i<=3*n; ++i) pre[i]=i; } int finds(int x){ return pre[x]=pre[x]==x?x:finds(pre[x]); } void unions(int x, int y){ int fx=finds(x), fy=finds(y); pre[fy]=fx; } int main(){ //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); cin>>n>>m; init(n); int t=0; for(int i=1; i<=m; ++i){ int t,x,y; scanf("%d%d%d",&t,&x,&y); if(x>n || y>n || x<1 || y<1 || (t==2 && x==y)){ printf("%d ",i); continue; } if(t==1){ if(finds(x)==finds(y+n) || finds(x)==finds(y+2*n)){ printf("%d ",i); } else{ unions(x,y); unions(x+n,y+n); unions(x+2*n,y+2*n); } } else{ if(finds(x)==finds(y) || finds(x)==finds(y+2*n)){ printf("%d ",i); } else{ unions(x,y+n); unions(x+n,y+2*n); unions(x+2*n,y); } } } return 0; }
原文:http://www.cnblogs.com/max88888888/p/7230782.html