一题:
检查读入优化
if (ch=='-')
少了一个等号
二题:LCA:SOJP942:Tree
#include<cctype>
#include<ctime>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
int root,n,m,cnt=0,lca,first[40002],f[40002][18],c[40002],deep[40002];
struct node{
int next,to;
}edge[40002];
inline void addedge(int x,int y){
cnt++;
edge[cnt].to=x;
edge[cnt].next=first[y];
first[y]=cnt;
}
inline void putroot(int root){
deep[root]=1;
c[1]=root;
for (int i=1,k=1;i<=k;i++){
for (int j=first[c[i]];j!=0;j=edge[j].next){
if(edge[j].to!=0){
deep[edge[j].to]=deep[c[i]]+1;
k++;
c[k]=edge[j].to;
}
}
}
}
inline int LCA(int x,int y){
if (deep[x]<deep[y])
swap(x,y);
int foot=deep[x]-deep[y];
for (int i=15;i>=0;i--)
if(foot>=(1<<i)){
x=f[x][i];
foot-=(1<<i);
}
if (x==y) return x;
for (int i=15;i>=0;i--)
if (f[x][i]!=f[y][i]){
x=f[x][i];
y=f[y][i];
}
return f[x][0];
}
int main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
cin>>n;
for (int i=1;i<=n;i++){
int x,y;
cin>>x>>y;
if (y==-1) root=x;
else f[x][0]=y;
addedge(x,y);
}
putroot(root);
for (int i=1;i<=15;i++)
for (int j=1;j<=40000;j++){
if (f[j][i-1]!=0)
f[j][i]=f[f[j][i-1]][i-1];
}
cin>>m;
for (int i=1;i<=m;i++){
int x,y;
cin>>x>>y;
lca=LCA(x,y);
if (x==y){
cout<<"0"<<endl;
continue;
}
if (lca==x){
cout<<"1"<<endl;
continue;
}
if (lca==y){
cout<<"2"<<endl;
continue;
}
cout<<"0"<<endl;
}
}
步骤:
STEP 1:单向建树
inline void addedge(int x,int y){
cnt++;
edge[cnt].to=x;
edge[cnt].next=first[y];
first[y]=cnt;
}
for (int i=1;i<=n;i++){
int x,y;
cin>>x>>y;
if (y!=-1) f[x][0]=y;
else root=-1;
addedge(x,y);
}
STEP 2:
从根开始往下标记深度:
inline void putroot(int x){
c[1]=x;
deep[x]=1;
for (int k=1,i=1;i<=k;i++){
for (int j=first[c[i]];j!=0;j=edge[j].next){
if (edge[j].to!=0){
deep[edge[j].to]=deep[c[i]]+1;
k++;
c[k]=edge[j].to;
}
}
}
}
STEP 3:初始化f[i][j]表示从i节点开始向上走2^j步所到达的值
注意顺序:必须是从小步向大步枚举,枚举每一步的时候把所有点枚举完
for (int j=1;j<=15;j++)//先枚举15且2^15>(40000/2):2^16一跳就跳出去了,没用
for (int i=1;i<=40000;i++)//再枚举40000
if (f[i][j-1])
f[i][j]=f[f[i][j-1]][j-1];
STEP 4:找每一对的LCA
inline int LCA(int x,int y){
if(deep[x]<deep[y])
swap(x,y);
int foot=deep[x]-deep[y];
for (int i=15;i>=0;i--)
if (foot>=(1<<i)){
foot-=(1<<i);
x=f[x][i];
}
if (x==y) return x;
for (int i=15;i>=0;i--){
if (f[x][i]!=f[y][i]){
x=f[x][i];
y=f[y][i];
}
}
return f[x][0];
}