8.13米哈游笔试(2/3)
T1
分别计算上下和左右移动的最小距离，取min即可
void solve(int u){
cin>>n>>m;
int a,b,c,d,e,f;
cin>>a>>b>>c>>d>>e>>f;
ll x=min(abs(a-c),n-abs(a-c)),y=min(abs(b-d),m-abs(b-d));
ll x1=min(abs(c-e),n-abs(c-e)),y1=min(abs(d-f),m-abs(d-f));
cout<<x1+x+y+y1<<endl;
}
T2
对于非叶子节点贡献+1
对于叶子节点，设叶子节点深度为d 贡献+(k-d+1)
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int>PII;
#define x first
#define y second
typedef long long ll;
const int N = 2E5 + 10, mod = 1e9 + 7;
int T, w[N], n,k;
vector<int>g[N];
int d[N];
ll res;
void dfs(int u,int fa,int d){
if(g[u].size()==1&&g[u][0]==fa){
if(d<=k){
res+=k-d+1;
}
return;
}
if(d<=k)res++;
for(int &x:g[u]){
if(x==fa)continue;
dfs(x,u,d+1);
}
}
void solve(int u) {
cin>>n>>k;
for(int i=1;i<n;i++){
int a,b;
cin>>a>>b;
g[a].push_back(b);
g[b].push_back(a);
}
dfs(1,-1,0);
cout<<res<<endl;
}
int main() {
T = 1;
for (int i = 1; i <= T; i++) {
solve(i);
}
return 0;
}
T3
期望DP 笔试的时候没推出来，纯菜呜呜
#米哈游笔试##秋招##后端开发##互联网大厂#
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