Codeforces Round #620 (Div. 2) A-F代码 （暂无记录题解）

A. Two Rabbits （手速题）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int t;
    cin>>t;
    while(t--){
        ll x,y,a,b;
        cin>>x>>y>>a>>b;
        ll sum = a+b;
        if((y-x)%sum == 0){
            cout<<(y-x)/sum<<endl;
        }
        else{
            cout<<-1<<endl;
        }
    }
    return 0;
}

B. Longest Palindrome （模拟 枚举）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int n,m;
    cin>>n>>m;
    vector<string> v;
    bool vis[n];
    memset(vis,0,sizeof(vis));
    for(int i = 0;i<n;i++){
        string s;
        cin>>s;
        v.push_back(s);
    }
    bool flag = false;
    string mid = "",ans = "";
    for(int i = 0;i<n;i++){
        string t = v[i];
        reverse(t.begin(),t.end());
        if(!flag && t == v[i]) {
            flag = true;
            mid = v[i];
            continue;
        }
        if(flag && t == v[i]) continue;
        for(int j = 0;j<n;j++){
            if(v[j] == t && vis[j] == 0){
                ans+=v[i];
                vis[i] = vis[j] = 1;
                break;
            }
        }        
    }
    string t = ans;
    reverse(t.begin(),t.end());
    ans+=mid,ans+=t;
    if(ans.length() == 0){
        cout<<0;return 0;
    }
    cout<<ans.length()<<endl;
    cout<<ans;
    return 0;
}

C. Air Conditioner （模拟 区间覆盖）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 105;
struct node{
    ll t,l,h;
    ll l1,r;
    bool operator < (node &x)const {
        return t<x.t;
    }
}c[maxn];
int main(){
    int q;
    cin>>q;
    while(q--){
        ll n,m;
        cin>>n>>m;
        for(int i = 1;i<=n;i++){
            cin>>c[i].t>>c[i].l>>c[i].h ; 
        } 
        sort(c+1,c+1+n);
        c[0].r = c[0].l1 = m; 
        bool flag = true;
        for(int i = 1;i<=n;i++){
            ll d = c[i].t - c[i-1].t;
            ll R = c[i-1].r + d;
            ll L = c[i-1].l1 - d;
            if(L>c[i].h || R< c[i].l){
                flag = false;
    //            cout<<"De"<<i<<endl;
                break;
            }
            c[i].r = min(R,c[i].h);
            c[i].l1 = max(L,c[i].l);
        }        
        if(flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}

D. Shortest and Longest LIS（贪心 构造）

#include<bits/stdc++.h>
using namespace std;
void solve(){
    int n;
    string s;
    cin>>n;
    cin>>s;
    int Min[n];
    int t = n;
    for(int i = 0;i<s.length();i++){
        int len = 0,indx = i;
        if(s[i] == '<'){
            while(s[i]=='<'){
                len++;
                i++;
            }
        }
        for(int j = i;j>=i-len;j--){
            Min[j] = t,t--;
        }
    } 
    if(t) Min[n-1] = t;
    t = 1;
    int Max[n];
    for(int i = 0;i<s.length();i++){
        int len = 0,indx = i;
        if(s[i] == '>'){
            while(s[i]=='>'){
                len++;
                i++;
            }
        }
        for(int j = i;j>=i-len;j--){
            Max[j] = t,t++;
        }
    }
    if(t == n) Max[n-1] = t;
    for(int i = 0;i<n;i++) cout<<Min[i]<<" ";
    cout<<endl;
    for(int i = 0;i<n;i++) cout<<Max[i]<<" ";
    cout<<endl; 
}
int main(){
    int t;
    cin>>t;
    while(t--){
        solve();
    } 
    return 0;
}

E. 1-Trees and Queries （最近公共祖先）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxbit = 20;
const int maxn = 1e5+5;
vector<int> G[maxn];
int depth[maxn];
int fa[maxn][maxbit];
int Log[maxn];
int N;
void pre(){
    Log[0] = -1;
    Log[1] = 0,Log[2] = 1;
    for(int i = 3;i<maxn;i++) Log[i] = Log[i/2] + 1;
} 
void dfs(int cur,int father){//dfs预处理 
    depth[cur] = depth[father] + 1;//当前结点的深度为父亲结点+1 
    fa[cur][0] = father;//更新当前结点的父亲结点 
    for(int j = 1;(1<<j)<=N;j++){//倍增更新当前结点的祖先 
        fa[cur][j] = fa[fa[cur][j-1]][j-1];
    }
    for(int i = 0;i<G[cur].size() ;i++){
        if(G[cur][i] != father) {//dfs遍历 
            dfs(G[cur][i],cur);
        }
    }
}
int LCA(int u,int v){
    if(depth[u]<depth[v]) swap(u,v);
    int dist = depth[u] - depth[v];//深度差 
    while(depth[u]!=depth[v]){//把较深的结点u倍增到与v高度相等 
        u = fa[u][Log[depth[u]-depth[v]]];
    }
    if(u == v) return u;//如果u倍增到v，说明v是u的LCA 
    for(int i = Log[depth[u]];i>=0;i--){//否则两者同时向上倍增 
        if(fa[u][i]!=fa[v][i]){//如果向上倍增的祖先不同，说明是可以继续倍增 
            u = fa[u][i];//替换两个结点 
            v = fa[v][i];
        }
    }
    return fa[u][0];//最终结果为u v向上一层就是LCA 
} 
int main()
{
    pre();
    cin>>N;
    for(int i = 1;i<N;i++){
        int u,v;
        cin>>u>>v;
        G[u].push_back(v),G[v].push_back(u);  
    }
    dfs(1,0);
    int q;cin>>q;
    while(q--){
        int x,y,a,b,k;
        cin>>x>>y>>a>>b>>k;
        int t1 = LCA(a,b),t2 = LCA(a,x),t3 = LCA(a,y),t4 = LCA(b,x),t5  = LCA(b,y);
        int dis = abs(depth[t1]-depth[a])+abs(depth[t1]-depth[b]);
        if(dis%2 == k%2 && dis <=k ){
//            cout<<"De1"<<endl;
            cout<<"YES"<<endl;
            continue;
        }
        int dis1 = abs(depth[t2]-depth[a])+abs(depth[t2]-depth[x]);
        int dis2 = abs(depth[t5]-depth[b])+abs(depth[t5]-depth[y]); 
        if( (dis1+dis2+1)%2 == k%2  && dis1+dis2+1 <= k){
//            cout<<"de2"<<endl;
            cout<<"YES"<<endl;
            continue;
        }
        dis1 = abs(depth[t3]-depth[a])+abs(depth[t3]-depth[y]);
        dis2 = abs(depth[t4]-depth[b])+abs(depth[t4]-depth[x]);
        if((dis1+dis2+1)%2 == k%2 && dis1+dis2+1 <= k){
//            cout<<"de3"<<endl;
            cout<<"YES"<<endl;
            continue;
        } 
        cout<<"NO"<<endl;
    }
    return 0;
}

F2. Animal Observation (hard version) （dp 滑动窗口 线段树区间更新最大值查询）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 4e4+5;
int dp[55][maxn];
int val[55][maxn];
int sum[55][maxn];
int n,m,k;
struct node{
    int l,r;
    int Max,lz;
}seg_t[maxn*4];
void build(int l,int r,int p){
    seg_t[p].l = l,seg_t[p].r = r;
    if(l == r) {
        seg_t[p].Max = 0;return ;
    }
    int mid = (l+r)>>1;
    build(l,mid,p*2);
    build(mid+1,r,p*2+1);
    seg_t[p].Max = max(seg_t[p*2].Max ,seg_t[p*2+1].Max );
}
void pushdown(int k){
    seg_t[k*2].lz +=seg_t[k].lz ;
    seg_t[k*2+1].lz +=seg_t[k].lz ;
    seg_t[k*2].Max +=seg_t[k].lz ;
    seg_t[k*2+1].Max +=seg_t[k].lz ;
    seg_t[k].lz = 0; 
}
void upd(int L,int R,int p,int v){
    if(seg_t[p].l == L && seg_t[p].r == R){
        seg_t[p].lz +=v;
        seg_t[p].Max +=v;
        return;
    }
    if(seg_t[p].lz ) pushdown(p);
    int mid = (seg_t[p].l + seg_t[p].r )>>1;
    if(R<=mid) upd(L,R,p*2,v);
    else if(L>mid) upd(L,R,p*2+1,v);
    else{
        upd(L,mid,p*2,v);
        upd(mid+1,R,p*2+1,v);
    }
    seg_t[p].Max = max(seg_t[p*2].Max ,seg_t[p*2+1].Max );
}
int query(int p,int L,int R){
    if(seg_t[p].l == L && seg_t[p].r == R) return seg_t[p].Max ;
    if(seg_t[p].lz ) pushdown(p);
    int mid = (seg_t[p].l +seg_t[p].r)>>1;
    if(R<=mid) return query(p*2,L,R);
    else if(L>mid) return query(p*2+1,L,R);
    else return max(query(p*2,L,mid),query(p*2+1,mid+1,R)); 
}
int main(){
    cin>>n>>m>>k;
    for(int i = 1;i<=n;i++){
        for(int j = 1;j<=m;j++) {
            cin>>val[i][j];
            sum[i][j] = sum[i][j-1] + val[i][j];
        }
    }
    for(int i = 1;i<=m-k+1;i++){
        dp[1][i] = sum[1][i+k-1] - sum[1][i-1]+sum[2][i+k-1]-sum[2][i-1];
    }
    for(int i = 2;i<=n;i++){
        memset(seg_t,0,sizeof(seg_t));
        build(1,2*m,1);
        for(int j = 1;j<=m;j++) upd(j,j,1,dp[i-1][j]);
        for(int j = 1;j<=k;j++) upd(1,j,1,-val[i][j]);
        for(int j = 1;j<=m-k+1;j++){
            dp[i][j] = max(dp[i][j],query(1,1,m)+sum[i][j+k-1]-sum[i][j-1]+sum[i+1][j+k-1]-sum[i+1][j-1]);
            upd(max(1,j-k+1),j,1,val[i][j]);
            upd(j+1,j+k,1,-val[i][j+k]);
        }
    } 
    int ans = 0;
    for(int i = 1;i<=m;i++) ans = max(ans,dp[n][i]);
    cout<<ans;
    return 0;
}