Codeforces Round #620 (Div. 2) A-F代码 （暂无记录题解）
A. Two Rabbits （手速题）
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int main(){ 5 int t; 6 cin>>t; 7 while(t--){ 8 ll x,y,a,b; 9 cin>>x>>y>>a>>b; 10 ll sum = a+b; 11 if((y-x)%sum == 0){ 12 cout<<(y-x)/sum<<endl; 13 } 14 else{ 15 cout<<-1<<endl; 16 } 17 } 18 return 0; 19 }
B. Longest Palindrome （模拟 枚举）
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int main(){ 5 int n,m; 6 cin>>n>>m; 7 vector<string> v; 8 bool vis[n]; 9 memset(vis,0,sizeof(vis)); 10 for(int i = 0;i<n;i++){ 11 string s; 12 cin>>s; 13 v.push_back(s); 14 } 15 bool flag = false; 16 string mid = "",ans = ""; 17 for(int i = 0;i<n;i++){ 18 string t = v[i]; 19 reverse(t.begin(),t.end()); 20 if(!flag && t == v[i]) { 21 flag = true; 22 mid = v[i]; 23 continue; 24 } 25 if(flag && t == v[i]) continue; 26 for(int j = 0;j<n;j++){ 27 if(v[j] == t && vis[j] == 0){ 28 ans+=v[i]; 29 vis[i] = vis[j] = 1; 30 break; 31 } 32 } 33 } 34 string t = ans; 35 reverse(t.begin(),t.end()); 36 ans+=mid,ans+=t; 37 if(ans.length() == 0){ 38 cout<<0;return 0; 39 } 40 cout<<ans.length()<<endl; 41 cout<<ans; 42 return 0; 43 }
C. Air Conditioner （模拟 区间覆盖）
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 105; 5 struct node{ 6 ll t,l,h; 7 ll l1,r; 8 bool operator < (node &x)const { 9 return t<x.t; 10 } 11 }c[maxn]; 12 int main(){ 13 int q; 14 cin>>q; 15 while(q--){ 16 ll n,m; 17 cin>>n>>m; 18 for(int i = 1;i<=n;i++){ 19 cin>>c[i].t>>c[i].l>>c[i].h ; 20 } 21 sort(c+1,c+1+n); 22 c[0].r = c[0].l1 = m; 23 bool flag = true; 24 for(int i = 1;i<=n;i++){ 25 ll d = c[i].t - c[i-1].t; 26 ll R = c[i-1].r + d; 27 ll L = c[i-1].l1 - d; 28 if(L>c[i].h || R< c[i].l){ 29 flag = false; 30 // cout<<"De"<<i<<endl; 31 break; 32 } 33 c[i].r = min(R,c[i].h); 34 c[i].l1 = max(L,c[i].l); 35 } 36 if(flag) cout<<"YES"<<endl; 37 else cout<<"NO"<<endl; 38 } 39 return 0; 40 }
D. Shortest and Longest LIS（贪心 构造）
1 #include<bits/stdc++.h> 2 using namespace std; 3 void solve(){ 4 int n; 5 string s; 6 cin>>n; 7 cin>>s; 8 int Min[n]; 9 int t = n; 10 for(int i = 0;i<s.length();i++){ 11 int len = 0,indx = i; 12 if(s[i] == '<'){ 13 while(s[i]=='<'){ 14 len++; 15 i++; 16 } 17 } 18 for(int j = i;j>=i-len;j--){ 19 Min[j] = t,t--; 20 } 21 } 22 if(t) Min[n-1] = t; 23 t = 1; 24 int Max[n]; 25 for(int i = 0;i<s.length();i++){ 26 int len = 0,indx = i; 27 if(s[i] == '>'){ 28 while(s[i]=='>'){ 29 len++; 30 i++; 31 } 32 } 33 for(int j = i;j>=i-len;j--){ 34 Max[j] = t,t++; 35 } 36 } 37 if(t == n) Max[n-1] = t; 38 for(int i = 0;i<n;i++) cout<<Min[i]<<" "; 39 cout<<endl; 40 for(int i = 0;i<n;i++) cout<<Max[i]<<" "; 41 cout<<endl; 42 } 43 int main(){ 44 int t; 45 cin>>t; 46 while(t--){ 47 solve(); 48 } 49 return 0; 50 }
E. 1-Trees and Queries （最近公共祖先）
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxbit = 20; 5 const int maxn = 1e5+5; 6 vector<int> G[maxn]; 7 int depth[maxn]; 8 int fa[maxn][maxbit]; 9 int Log[maxn]; 10 int N; 11 void pre(){ 12 Log[0] = -1; 13 Log[1] = 0,Log[2] = 1; 14 for(int i = 3;i<maxn;i++) Log[i] = Log[i/2] + 1; 15 } 16 void dfs(int cur,int father){//dfs预处理 17 depth[cur] = depth[father] + 1;//当前结点的深度为父亲结点+1 18 fa[cur][0] = father;//更新当前结点的父亲结点 19 for(int j = 1;(1<<j)<=N;j++){//倍增更新当前结点的祖先 20 fa[cur][j] = fa[fa[cur][j-1]][j-1]; 21 } 22 for(int i = 0;i<G[cur].size() ;i++){ 23 if(G[cur][i] != father) {//dfs遍历 24 dfs(G[cur][i],cur); 25 } 26 } 27 } 28 int LCA(int u,int v){ 29 if(depth[u]<depth[v]) swap(u,v); 30 int dist = depth[u] - depth[v];//深度差 31 while(depth[u]!=depth[v]){//把较深的结点u倍增到与v高度相等 32 u = fa[u][Log[depth[u]-depth[v]]]; 33 } 34 if(u == v) return u;//如果u倍增到v，说明v是u的LCA 35 for(int i = Log[depth[u]];i>=0;i--){//否则两者同时向上倍增 36 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 37 u = fa[u][i];//替换两个结点 38 v = fa[v][i]; 39 } 40 } 41 return fa[u][0];//最终结果为u v向上一层就是LCA 42 } 43 int main() 44 { 45 pre(); 46 cin>>N; 47 for(int i = 1;i<N;i++){ 48 int u,v; 49 cin>>u>>v; 50 G[u].push_back(v),G[v].push_back(u); 51 } 52 dfs(1,0); 53 int q;cin>>q; 54 while(q--){ 55 int x,y,a,b,k; 56 cin>>x>>y>>a>>b>>k; 57 int t1 = LCA(a,b),t2 = LCA(a,x),t3 = LCA(a,y),t4 = LCA(b,x),t5 = LCA(b,y); 58 int dis = abs(depth[t1]-depth[a])+abs(depth[t1]-depth[b]); 59 if(dis%2 == k%2 && dis <=k ){ 60 // cout<<"De1"<<endl; 61 cout<<"YES"<<endl; 62 continue; 63 } 64 int dis1 = abs(depth[t2]-depth[a])+abs(depth[t2]-depth[x]); 65 int dis2 = abs(depth[t5]-depth[b])+abs(depth[t5]-depth[y]); 66 if( (dis1+dis2+1)%2 == k%2 && dis1+dis2+1 <= k){ 67 // cout<<"de2"<<endl; 68 cout<<"YES"<<endl; 69 continue; 70 } 71 dis1 = abs(depth[t3]-depth[a])+abs(depth[t3]-depth[y]); 72 dis2 = abs(depth[t4]-depth[b])+abs(depth[t4]-depth[x]); 73 if((dis1+dis2+1)%2 == k%2 && dis1+dis2+1 <= k){ 74 // cout<<"de3"<<endl; 75 cout<<"YES"<<endl; 76 continue; 77 } 78 cout<<"NO"<<endl; 79 } 80 return 0; 81 }
F2. Animal Observation (hard version) （dp 滑动窗口 线段树区间更新最大值查询）
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 4e4+5; 4 int dp[55][maxn]; 5 int val[55][maxn]; 6 int sum[55][maxn]; 7 int n,m,k; 8 struct node{ 9 int l,r; 10 int Max,lz; 11 }seg_t[maxn*4]; 12 void build(int l,int r,int p){ 13 seg_t[p].l = l,seg_t[p].r = r; 14 if(l == r) { 15 seg_t[p].Max = 0;return ; 16 } 17 int mid = (l+r)>>1; 18 build(l,mid,p*2); 19 build(mid+1,r,p*2+1); 20 seg_t[p].Max = max(seg_t[p*2].Max ,seg_t[p*2+1].Max ); 21 } 22 void pushdown(int k){ 23 seg_t[k*2].lz +=seg_t[k].lz ; 24 seg_t[k*2+1].lz +=seg_t[k].lz ; 25 seg_t[k*2].Max +=seg_t[k].lz ; 26 seg_t[k*2+1].Max +=seg_t[k].lz ; 27 seg_t[k].lz = 0; 28 } 29 void upd(int L,int R,int p,int v){ 30 if(seg_t[p].l == L && seg_t[p].r == R){ 31 seg_t[p].lz +=v; 32 seg_t[p].Max +=v; 33 return; 34 } 35 if(seg_t[p].lz ) pushdown(p); 36 int mid = (seg_t[p].l + seg_t[p].r )>>1; 37 if(R<=mid) upd(L,R,p*2,v); 38 else if(L>mid) upd(L,R,p*2+1,v); 39 else{ 40 upd(L,mid,p*2,v); 41 upd(mid+1,R,p*2+1,v); 42 } 43 seg_t[p].Max = max(seg_t[p*2].Max ,seg_t[p*2+1].Max ); 44 } 45 int query(int p,int L,int R){ 46 if(seg_t[p].l == L && seg_t[p].r == R) return seg_t[p].Max ; 47 if(seg_t[p].lz ) pushdown(p); 48 int mid = (seg_t[p].l +seg_t[p].r)>>1; 49 if(R<=mid) return query(p*2,L,R); 50 else if(L>mid) return query(p*2+1,L,R); 51 else return max(query(p*2,L,mid),query(p*2+1,mid+1,R)); 52 } 53 int main(){ 54 cin>>n>>m>>k; 55 for(int i = 1;i<=n;i++){ 56 for(int j = 1;j<=m;j++) { 57 cin>>val[i][j]; 58 sum[i][j] = sum[i][j-1] + val[i][j]; 59 } 60 } 61 for(int i = 1;i<=m-k+1;i++){ 62 dp[1][i] = sum[1][i+k-1] - sum[1][i-1]+sum[2][i+k-1]-sum[2][i-1]; 63 } 64 for(int i = 2;i<=n;i++){ 65 memset(seg_t,0,sizeof(seg_t)); 66 build(1,2*m,1); 67 for(int j = 1;j<=m;j++) upd(j,j,1,dp[i-1][j]); 68 for(int j = 1;j<=k;j++) upd(1,j,1,-val[i][j]); 69 for(int j = 1;j<=m-k+1;j++){ 70 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]); 71 upd(max(1,j-k+1),j,1,val[i][j]); 72 upd(j+1,j+k,1,-val[i][j+k]); 73 } 74 } 75 int ans = 0; 76 for(int i = 1;i<=m;i++) ans = max(ans,dp[n][i]); 77 cout<<ans; 78 return 0; 79 }
