一道带权并查集题目.
带权并查集的重点是信息的合并.
这类题出现得并不多,练习一下.
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define FILE "dealing" 5 #define up(i,j,n) for(int i=j;i<=n;i++) 6 #define db long double 7 #define pii pair<int,int> 8 #define pb push_back 9 #define mem(a,L) memset(a,0,sizeof(int)*(L+1)) 10 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 11 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 12 template<class T> inline T squ(T a){return a*a;} 13 const ll maxn=210000+10,inf=1e9+10,mod=10000007,M=9988440; 14 ll read(){ 15 ll x=0,f=1,ch=getchar(); 16 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 17 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 18 return x*f; 19 } 20 int l[maxn],r[maxn],fa[maxn]; 21 int getfa(int x){ 22 if(fa[x]==x)return x; 23 int y=getfa(fa[x]); 24 if(y==fa[x])return y; 25 l[x]+=l[fa[x]],r[x]+=r[fa[x]]; 26 fa[x]=y; 27 return y; 28 } 29 int n,m,Q; 30 int f1[maxn],f2[maxn],d[maxn],op[maxn]; 31 int xx[maxn],yy[maxn],t[maxn],ans[maxn]; 32 vector<int> linkk[maxn]; 33 int main(){ 34 freopen(FILE".in","r",stdin); 35 freopen(FILE".out","w",stdout); 36 n=read(),m=read(); 37 char ch; 38 up(i,1,m){ 39 f1[i]=read(),f2[i]=read(),d[i]=read(); 40 scanf(" %c",&ch); 41 if(ch=='E')op[i]=1; 42 else if(ch=='W')op[i]=2; 43 else if(ch=='N')op[i]=3; 44 else if(ch=='S')op[i]=4; 45 } 46 Q=read(); 47 up(i,1,Q){ 48 xx[i]=read(),yy[i]=read(),t[i]=read(); 49 linkk[t[i]].push_back(i); 50 } 51 up(i,1,n)fa[i]=i; 52 up(i,1,m){ 53 int x=f1[i],y=f2[i],fx=getfa(x),fy=getfa(y); 54 if(fx!=fy){ 55 if(fx!=x)l[fx]=-l[x],r[fx]=-r[x],l[x]=r[x]=0,fa[fx]=x,fa[x]=x; 56 if(fy!=y)l[fy]=-l[y],r[fy]=-r[y],l[y]=r[y]=0,fa[fy]=y,fa[y]=y; 57 fa[x]=y; 58 if(op[i]==1)l[x]=d[i]; 59 else if(op[i]==2)l[x]=-d[i]; 60 else if(op[i]==3)r[x]=d[i]; 61 else r[x]=-d[i]; 62 } 63 for(int j=0;j<linkk[i].size();j++){ 64 int t=linkk[i][j]; 65 int x=xx[t],y=yy[t]; 66 int fx=getfa(x),fy=getfa(y); 67 if(fx!=fy)ans[t]=-1; 68 else ans[t]=abs(l[x]-l[y])+abs(r[x]-r[y]); 69 } 70 } 71 up(i,1,Q)printf("%d ",ans[i]); 72 return 0; 73 }