1、二叉树的最大路径和;66%
#include<bits/stdc++.h>
using namespace std;
struct node{
int val;
vector<node*> childs;
node(int val): val(val){};
node(): val(0){};
};
long ret = INT_MIN;
long get_max_path(node* n){
if(n == NULL)
return 0;
if(n->childs.size() == 0)
return n->val;
long cur_path_len = 0;
long left = 0, right = 0;
left = get_max_path(n->childs[0]);
if(n->childs.size()> 1)
right = get_max_path(n->childs[1]);
cur_path_len = left + right+ (long)n->val;
cout << cur_path_len << ' ' << n->val << endl;
ret = max(ret, cur_path_len);
return max(left, right) + (long)n->val;
}
int main(){
int n;
cin >> n;
vector<node> e(n);
for(int i=0; i< n; i++){
cin >> e[i].val;
}
for(int i=0; i< n; i++){
if(i == 0)
continue;
int father;
cin >> father;
e[father].childs.push_back(&e[i]);
}
for(int i=0; i< n; i++){
cout << e[i].childs.size() << endl;
}
node* root = &e[0];
get_max_path(root);
cout << ret;
return 0;
}
2、能到达出口的最小操作次数;100%
#include<bits/stdc++.h>
using namespace std;
bool is_location(int x, int y, int m, int n){
return x>=0 && x< m && y>= 0 && y< n;
}
int main(){
int m, n;
pair<int, int> start, end;
cin>> m >> n;
cin >> start.first >> start.second;
cin >> end.first >> end.second;
start.first -= 1; start.second -= 1;
end.first -= 1; end.second -= 1;
vector<string> maps(m);
for(int i=0; i< m; i++){
cin >> maps[i];
}
// 节点是否访问过
unordered_set<int> visited;
visited.insert(start.first*n + start.second);
queue<pair<int, int>> q;
q.push(start);
long ret = 0;
vector<pair<int, int>> D = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
while(!q.empty()){
int size = q.size();
ret += 1;
for(int i=0; i< size; i++){
auto front = q.front(); q.pop();
for(auto d: D){
pair<int, int> next = {front.first+d.first, front.second+d.second};
if(is_location(next.first, next.second, m, n)){
if(visited.find(next.first*n+next.second) != visited.end() || maps[next.first][next.second] == '*')
continue;
if(next.first == end.first && next.second == end.second){
cout << ret;
return 0;
}
q.push(next);
visited.insert(next.first*n+next.second);
}
}
}
}
cout << -1;
return 0;
}
3、使用Bash写出0-n中所有7的倍数;
题目量太大了!!!一共一个半小时
1、视频题,说你在平常时间怎么做出重大的决定;
2、选择题20题,有C++问题,还有一些推理题
3、解答题:5题
4、编程题:3题;这个第一题的二叉树他用ACM模式说,并且自己的输入他都没有说清楚,结果搞了很多时间;
#S2C笔试#