20140711 eat

不多说 NOI2001食物链

 

核心就是并查集，并查集数组中存0 1 2 3 表示未分营养级 A  B  C

 然后再一个个恶心的判断....

inline bool Union(int x,int y,int D){
int a=find(x), b=find(y);
if(a==b){
if(D==1&&rank[x]!=rank[y]) return false;
if(D==2){
if(rank[x]==2&&rank[y]!=1)return false;
if(rank[x]==1&&rank[y]!=0)return false;
if(rank[x]==0&&rank[y]!=2)return false;
}
return true;
}
f[a] = b;
if(D==2){
rank[a] = (rank[y]-rank[x]+3+1)%3;
}
else{
rank[a] = (rank[y]-rank[x]+3)%3;
}
return true;
}

 

其实一开始我想的是用3个set，原因嘛，跟set（那道题）一样........

 

#include<cstdio> 
 
const int N = 100005; 
int n, m, f[N], rank[N]; 
 
inline void init(){ 
    for(int i=0; i<=n; ++i) 
        f[i]=i,rank[i]=0; 
} 
 
int find(int x){ 
    if(x==f[x])return x; 
    int fa=f[x]; 
    f[x]=find(f[x]); 
    rank[x] = (rank[x]+rank[fa])%3; 
    return f[x]; 
} 
 
inline bool Union(int x,int y,int D){ 
    int a=find(x), b=find(y); 
    if(a==b){ 
        if(D==1&&rank[x]!=rank[y]) return false; 
        if(D==2){ 
            if(rank[x]==2&&rank[y]!=1)return false; 
            if(rank[x]==1&&rank[y]!=0)return false; 
            if(rank[x]==0&&rank[y]!=2)return false; 
        } 
        return true; 
    } 
    f[a] = b; 
    if(D==2){ 
        rank[a] = (rank[y]-rank[x]+3+1)%3; 
    } 
    else{ 
        rank[a] = (rank[y]-rank[x]+3)%3;   
    } 
    return true; 
} 
 
int main(){ 
    freopen("eat.in","r",stdin);
    freopen("eat.out","w",stdout);
    int x,y,d; 
    scanf("%d%d",&n,&m); 
    init(); 
    int cnt=0; 
    for(int i=0; i<m; ++i){ 
        scanf("%d%d%d",&d,&x,&y); 
        if(x>n || y>n || d==2&&x==y){ 
            ++cnt; continue; 
        } 
        if(!Union(x,y,d)) { 
            ++cnt; 
        } 
    } 
    printf("%d\n",m-cnt); 
    return 0; 
}