第五章树25 AcWing 1644. 二叉树中的最低公共祖先

Myth丶恋晨 2023-09-24 14:07 68阅读 0赞

## 第五章 树25 AcWing 1644. 二叉树中的最低公共祖先 ##

### 原题链接 ###

[AcWing 1644. 二叉树中的最低公共祖先][AcWing 1644.]

### 算法标签 ###

LCA

### 思路 ###

与该题极为相似， 具体思路可见该题解[第五章 树 24 AcWing 1636. 最低公共祖先][_ 24 AcWing 1636.]

### 代码 ###

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #include<bits/stdc++.h>
    #define int long long
    #define x first
    #define y second
    #define ump unordered_map
    #define pq priority_queue
    #define rep(i, a, b) for(int i=a;i<b;++i)
    #define Rep(i, a, b) for(int i=a;i>=b;--i)
    using namespace std;
    typedef pair<int, int> PII;
    const int N = 10005;
    //int t, n, m, cnt, ans; 
    int n, m, in[N], pre[N], seq[N], f[N], d[N];
    ump<int, int> pos;
    inline int rd(){
       int s=0,w=1;
       char ch=getchar();
       while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
       while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
       return s*w;
    }
    void put(int x) {
        if(x<0) putchar('-'),x=-x;
        if(x>=10) put(x/10);
        putchar(x%10^48);
    }
    int build(int il, int ir, int pl, int pr, int dd){
        int root=pre[pl];
        int k=root;
        d[root]=dd;
        if(il<k){
            f[build(il, k-1, pl+1, k-1-il+pl+1, dd+1)]=root;
        }
        if(ir>k){
            f[build(k+1, ir, k-1-il+pl+1+1, pr-1, dd+1)]=root;
        }
        return root;
    }
    
    signed main(){
    	ios::sync_with_stdio(false);
    	cin.tie(0);
    	cout.tie(0);
    	rep(i, 0, N){
    	    seq[i]=rd();
    	    pos[seq[i]]=i;
    	    in[i]=i;
    	}
    	rep(i, 0, N){
    	    pre[i]=rd();
    	    pre[i]=pos[pre[i]];
    	}
    	build(0, n-1, 0, n-1, 0);
    	while(m--){
    	    int u=rd(), v=rd();
    	    int uu=u, vv=v;
    	    if(pos.count(u)&&pos.count(v)){
    	        u=pos[u], v=pos[v];
    	        while(u!=v){
    	            if(d[u]>d[v]){
    	                u=f[u];
    	            }else if(d[u]<d[v]){
    	                v=f[v];
    	            }else{
    	                u=f[u],v=f[v];
    	            }
    	        }
    	        if(uu!=u&&vv!=v){
    	            printf("LCA of %lld and %lld is %lld.\n", seq[uu], seq[vv], seq[u]);
    	        }else if(uu=u){
    	            printf("%lld is an ancestor of %lld.\n", seq[uu], seq[u]);
    	        }else if(vv=v){
    	            printf("%lld is an ancestor of %lld.\n", seq[vv], seq[v]);
    	        }
    	    }else{
    	        if(pos.count(u)==0&&pos.count(v)==0){
    	            printf("ERROR: %lld and %lld are not found.", seq[uu], seq[vv]);  
    	        }else if(pos.count(u)==0){
    	            printf("ERROR: %lld is not found.", seq[uu]);
    	        }else if(pos.count(v)==0){
    	            printf("ERROR: %lld is not found.", seq[vv]);
    	        }
    	    }
    	}
    	return 0;
    }

### 参考文献 ###

[AcWing 1644. 二叉树中的最低公共祖先（PAT甲级辅导课）y总视频讲解][AcWing 1644. _PAT_y]

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![在这里插入图片描述][9cb4cb80a18345158f8a69e2f196e5b9.png]

[AcWing 1644.]: https://www.acwing.com/problem/content/1646/
[_ 24 AcWing 1636.]: https://blog.csdn.net/T_Y_F_/article/details/127826600
[AcWing 1644. _PAT_y]: https://www.acwing.com/video/1200/
[9cb4cb80a18345158f8a69e2f196e5b9.png]: https://img-blog.csdnimg.cn/9cb4cb80a18345158f8a69e2f196e5b9.png