【专题复习1：树的遍历】1086、1053、1079、1090、1094、1106、1119、1138、1143、1151
1086
1086
发现一个问题，如果vector事先resize了的话，再用push_back()添加元素就会失败，这种情况应该像数组那样处理。All in all，要么不resize然后push_back()；要么resize用下标添加。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
struct node
{
int data;
node* lc;
node* rc;
};
vector<int> pre,in;
int n;
node* create(int il,int ir,int pl,int pr)
{
if(pl>pr) return nullptr;
node* root=new node;
root->data=pre[pl];
int k=0;
while(in[k]!=pre[pl]) k++;
root->lc=create(il,k-1,pl+1,pl+k-il);
root->rc=create(k+1,ir,pl+k-il+1,pr);
return root;
}
int num=0;
void postOrder(node* root)
{
if(root==nullptr) return;
postOrder(root->lc);
postOrder(root->rc);
cout<<root->data;
num++;
if(num<n) cout<<" ";
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
string s;
int x,inx=0,index=0;
stack<int> sk;
cin>>n;
//pre.resize(n);
in.resize(n);
for(int i=0;i<2*n;i++){
cin>>s;
if(s=="Push"){
cin>>x;
//pre[inx++]=x;
pre.push_back(x);
sk.push(x);
}else{
in[index++]=sk.top();
sk.pop();
}
}
node* root=create(0,n-1,0,n-1);
postOrder(root);
return 0;
}
1053
1053
update：如果子节点有相同值时原先的算法将会出错，参见
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#include <cstdio>
#include <iostream> // 不知道为什么不用这个头文件PAT会对greater报错
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 110;
struct node {
int weight;
vector<int> child;
} Node[maxn];
int n, m, s;
vector<int> path; //路径
vector<vector<int>> ans; //存放所有路径
void DFS(int index, vector<int> path, int sum) {
if (sum > s) return;
if (sum == s) {
if (Node[index].child.size() != 0) return; //还未到叶子结点
ans.push_back(path);
return;
}
for (int i = 0; i < Node[index].child.size(); i++) {
int child = Node[index].child[i];
path.push_back(Node[child].weight);
DFS(child, path, sum + Node[child].weight);
path.pop_back();
}
}
int main() {
scanf("%d %d %d", &n, &m, &s);
for (int i = 0; i < n; i++) {
scanf("%d", &Node[i].weight);
}
int id, k, child;
for (int i = 0; i < m; i++) {
scanf("%d %d", &id, &k);
for (int j = 0; j < k; j++) {
scanf("%d", &child);
Node[id].child.push_back(child);
}
}
path.push_back(Node[0].weight);
DFS(0, path, Node[0].weight);
//对每种情况按字典序从大到小排序
sort(ans.begin(), ans.end(), greater<vector<int>>());
for (int i = 0; i < ans.size(); i++) {
printf("%d", ans[i][0]);
for (int j = 1; j < ans[i].size(); j++) {
printf(" %d", ans[i][j]);
}
printf("\n");
}
return 0;
}
1079
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#include <bits/stdc++.h>
using namespace std;
struct node
{
int data;
vector<int> child;
};
vector<node> v;
int n;
double p,r,ans;
void dfs(int index,int depth)
{
if(v[index].child.size()==0){
ans+=v[index].data*pow(1+r,depth);
return ;
}
for(int i=0;i<v[index].child.size();i++)
dfs(v[index].child[i],depth+1);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int num,c;
cin>>n>>p>>r;
r/=100;
v.resize(n);
for(int i=0;i<n;i++){
cin>>num;
if(num==0) cin>>v[i].data;
else{
for(int j=0;j<num;j++){
cin>>c;
v[i].child.push_back(c);
}
}
}
dfs(0,0);
printf("%.1f\n",ans*p);
return 0;
}
1090
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#include <bits/stdc++.h>
using namespace std;
vector<int> v[100010];
int n,root,maxnum,maxdepth;
double p,r;
void dfs(int index,int depth)
{
if(v[index].size()==0){
if(maxdepth==depth) maxnum++;
else if(depth>maxdepth){
maxdepth=depth;
maxnum=1;
}
return ;
}
for(int i=0;i<v[index].size();i++)
dfs(v[index][i],depth+1);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int temp;
cin>>n>>p>>r;
r/=100;
for(int i=0;i<n;i++){
cin>>temp;
if(temp!=-1)
v[temp].push_back(i);
else root=i;
}
dfs(root,0);
printf("%.2f %d\n",p*pow(1+r,maxdepth),maxnum);
return 0;
}
1094
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#include <bits/stdc++.h>
using namespace std;
vector<int> v[100010];
int n,m;
int book[100];
void dfs(int index,int depth)
{
book[depth]++;
for(int i=0;i<v[index].size();i++)
dfs(v[index][i],depth+1);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int index,num,temp;
cin>>n>>m;
for(int i=0;i<m;i++){
cin>>index>>num;
for(int j=0;j<num;j++){
cin>>temp;
v[index].push_back(temp);
}
}
dfs(1,1);
int maxnum=0,maxdepth=0;
for(int i=1;i<100;i++){
if(book[i]>maxnum){
maxdepth=i;
maxnum=book[i];
}
}
cout<<maxnum<<" "<<maxdepth;
return 0;
}
1106
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#include <bits/stdc++.h>
using namespace std;
vector<int> v[100010];
int n;
double p,r;
int minnum=1,mindepth=100000000;
void dfs(int index,int depth)
{
if(mindepth<depth) return;
if(v[index].size()==0){
if(depth<mindepth){
mindepth=depth;
minnum=1;
}else if(depth==mindepth)
minnum++;
}
for(int i=0;i<v[index].size();i++)
dfs(v[index][i],depth+1);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int num,temp;
cin>>n>>p>>r;
for(int i=0;i<n;i++){
cin>>num;
for(int j=0;j<num;j++){
cin>>temp;
v[i].push_back(temp);
}
}
dfs(0,0);
printf("%.4f %d",p*pow(1+r/100,mindepth),minnum);
return 0;
}
1119
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#include <bits/stdc++.h>
using namespace std;
vector<int> in,pre,post;
bool flag=true;
void getIn(int prel,int prer,int postl,int postr)
{
if(prel==prer){
in.push_back(pre[prel]);
return;
}
if(pre[prel]==post[postr]){
int i=prel+1;
while(i<=prer&&pre[i]!=post[postr-1]) i++;
if(i-prel>1)
getIn(prel+1,i-1,postl,postl+i-prel-1-1);
else
flag=false;
in.push_back(post[postr]);
getIn(i,prer,postl+i-prel-1,postr-1);
}
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int n;
cin>>n;
pre.resize(n);post.resize(n);
for(int i=0;i<n;i++) cin>>pre[i];
for(int i=0;i<n;i++) cin>>post[i];
getIn(0,n-1,0,n-1);
printf("%s\n%d",flag==true?"Yes":"No",in[0]);
for(int i=1;i<n;i++)
printf(" %d",in[i]);
printf("\n");
return 0;
}
1138
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#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct Node {
int val;
Node *left, *right;
Node(int data) : val(data), left(NULL), right(NULL) {};
};
vector<int> in, pre;
bool flag = 0;
Node *creat(int prel, int prer, int inl, int inr) {
if (prel > prer)return NULL;
Node *root = new Node(pre[prel]);
int k;
for (k = inl; k <= inr && in[k] != pre[prel]; k++);
int numleft = k - inl;
root->left = creat(prel + 1, prel + numleft, inl, k - 1);
root->right = creat(prel + numleft + 1, prer, k + 1, inr);
return root;
}
void post(Node *root) {
if (root == NULL)return;
post(root->left);
post(root->right);
if (flag)return;
printf("%d", root->val);
flag = true;
}
int main() {
int n;
scanf("%d", &n);
in.resize(n);
pre.resize(n);
for (int i = 0; i < n; i++)scanf("%d", &pre[i]);
for (int i = 0; i < n; i++)scanf("%d", &in[i]);
Node *root = creat(0, n - 1, 0, n - 1);
post(root);
return 0;
}
1143
1143
还是不太理解。。。
尝试理解这段话==
由于，二叉排序树性质，最近祖先的值一定是介于两者之间的.
根据前序遍历顺序来找节点的原因是：
前序遍历的特点是根、左子树、右子树。根据前序遍历的顺序，对于当前根来说：
1.如果a,b分布在当前根的两侧或者根是a|b,那么答案就找到
2.如果a,b分布在当前根的左侧，由于根左右，下一个点就是左子树的根，顺理成章
3.如果a,b分布在当前根的右侧，根左右，整个左子树都小于a,b,都会不出结果,等到遍历到右子树就行
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#include <bits/stdc++.h>
using namespace std;
map<int,bool> mp;
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int m,n,u,v,a;
cin>>m>>n;
vector<int> pre(n);
for(int i=0;i<n;i++){
cin>>pre[i];
mp[pre[i]]=true;
}
for(int i=0;i<m;i++){
cin>>u>>v;
for(int j=0;j<n;j++){
a=pre[j];
if(a>u&&a<v||a>v&&a<u||a==u||a==v) break;
}
if(mp[u]==false&&mp[v]==false)
printf("ERROR: %d and %d are not found.\n", u, v);
else if(mp[u]==false||mp[v]==false)
printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
else if(a==u||a==v)
printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
else
printf("LCA of %d and %d is %d.\n", u, v, a);
}
return 0;
}
1151
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#include <bits/stdc++.h>
using namespace std;
map<int,int> pos;
vector<int> in,pre;
void lca(int inl,int inr,int preRoot,int a,int b)
{
if(inl>inr) return;
int inRoot=pos[pre[preRoot]],aIn=pos[a],bIn=pos[b];
if(aIn<inRoot&&bIn<inRoot)
lca(inl,inRoot-1,preRoot+1,a,b);
else if((aIn<inRoot&&bIn>inRoot)||(aIn>inRoot&&bIn<inRoot))
printf("LCA of %d and %d is %d.\n", a, b, in[inRoot]);
else if(aIn>inRoot&&bIn>inRoot)
lca(inRoot+1,inr,preRoot+1+(inRoot-inl),a,b);
else if(aIn==inRoot)
printf("%d is an ancestor of %d.\n", a, b);
else if(bIn==inRoot)
printf("%d is an ancestor of %d.\n", b, a);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int m,n,a,b;
cin>>m>>n;
in.resize(n+1);pre.resize(n+1);
for(int i=1;i<=n;i++){
cin>>in[i];
pos[in[i]]=i;
}
for(int i=1;i<=n;i++) cin>>pre[i];
for(int i=0;i<m;i++){
cin>>a>>b;
if(pos[a]==0&&pos[b]==0)
printf("ERROR: %d and %d are not found.\n", a, b);
else if(pos[a]==0||pos[b]==0)
printf("ERROR: %d is not found.\n", pos[a] == 0 ? a : b);
else
lca(1,n,1,a,b);
}
return 0;
}
本文来自博客园，作者：勇往直前的力量，转载请注明原文链接：https://www.cnblogs.com/moonlight1999/p/15938875.html
