2020 ICPC 银川区域赛 ABEGJK
A. Best Player
模拟。
题意是分别站在 \(X,Y,Z\) 轴上看另外两维构成的点，谁看到的更多。
分别对三维维护一个 \(\text{pair}\) 去重比较即可。
点击查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define ll long long
const int N=5003;
#define int long long
int id=0;
void solve() {
int n;
cin >> n;
set<array<int,2>>stx,sty,stz;
int ans = 0;
char res;
for (int i = 0; i < n; ++i) {
int x, y, z;
cin >> x >> y >> z;
stx.insert({y,z});
sty.insert({x,z});
stz.insert({x,y});
}
if(stx.size()>ans){
ans=stx.size();
res='X';
}
if(sty.size()>ans){
ans=sty.size();
res='Y';
}
if(stz.size()>ans){
ans=stz.size();
res='Z';
}
cout << res << endl;
}
signed main() {
ios::sync_with_stdio(false), cin.tie(0);
int t = 1;
// cin >> t;
while (t--) {
solve();
}
return 0;
}
B. The Great Wall
dp。
设 \(dp_{i,j,0/1/2/3}\) 表示第 \(i\) 个数被分到 \(j\) 组，且 未确定最大最小值 / 已确定最小值 / 已确定最大值 / 已确定最大最小值 的最大答案。
那么当 \(a_i\) 不产生贡献 / 作为最小值 / 作为最大值 时有转移方程为：
\[dp_{i,j,0} = dp_{i,j-1,3}\\
dp_{i,j,1} = \max(dp_{i-1,j,1},dp_{i-1,j-1,3}-a_i,dp_{i-1,j,0}-a_i)\\
dp_{i,j,2} = \max(dp_{i-1,j,2},dp_{i-1,j-1,3}+a_i,dp_{i-1,j,0}+a_i)\\
dp_{i,j,3} = \max(dp_{i-1,j,3},dp_{i-1,j-1,3},dp_{i-1,j-1,1}+a_i,dp_{i-1,j,2}-a_i)\\
\]
\(4\times 10^8\) 的空间存不下，需要滚动。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<int> a(n + 1);
for(int i = 1; i <= n; i += 1) {
cin >> a[i];
}
const i64 inf = -1E9;
vector<array<i64,4>> dp(n + 1, {inf, inf, inf, inf});
auto ndp = dp;
dp[0][3] = 0;
for(int i = 1; i <= n; i += 1) {
for(int j = 1; j <= i; j += 1) {
ndp[j][0] = dp[j - 1][3];
ndp[j][1] = max({dp[j][1], dp[j - 1][3] - a[i], dp[j][0] - a[i]});
ndp[j][2] = max({dp[j][2], dp[j - 1][3] + a[i], dp[j][0] + a[i]});
ndp[j][3] = max({dp[j][3], dp[j - 1][3], dp[j][1] + a[i], dp[j][2] - a[i]});
}
swap(dp, ndp);
}
for(int i = 1; i <= n; i += 1) {
cout << dp[i][3] << "\n";
}
return 0;
}
E. Isomerism
模拟。
读懂题就会做了，直接看代码即可。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
void solve() {
map<string,int> mp;
mp["-F"] = 8, mp["-Cl"] = 7, mp["-Br"] = 6, mp["-I"] = 5;
mp["-CH3"] = 4, mp["-CH2CH3"] = 3, mp["-CH2CH2CH3"] = 2, mp["-H"] = 1;
string R[4];
cin >> R[0] >> R[1] >> R[2] >> R[3];
if(R[0] == R[2] || R[1] == R[3]){
cout << "None\n";
return;
}
if(R[0] == R[1] || R[2] == R[3]){
cout << "Cis\n";
return;
}
if(R[0] == R[3] || R[2] == R[1]){
cout << "Trans\n";
return;
}
if(mp[R[0]] > mp[R[2]] && mp[R[1]] > mp[R[3]] || mp[R[0]] < mp[R[2]] && mp[R[1]] < mp[R[3]]){
cout << "Zasamman\n";
return;
}
cout << "Entgegen\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
G. Photograph
解法一：链表。
模拟一个 \(1\sim n\) 的链表，每次按顺序加进来一个数反过来看就是每次从链表中删掉一个数，删掉这个数后左右两边的指针更新一下即可，先算出总的贡献，删掉数后再减掉那个数的贡献就能得到当前序列的贡献。
点击查看代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, q;
cin >> n >> q;
vector<int> a(n + 1);
for(int i = 1;i <= n;i += 1){
cin >> a[i];
}
vector<int> p(n + 1);
for(int i = 1;i <= n;i += 1){
cin >> p[i];
}
i64 ans = 0;
auto query = [&]()->void{
i64 sum = 0;
vector<int> l(n + 1, -1), r(n + 1, -1);
for(int i = 2;i <= n;i += 1){
l[i] = i - 1;
r[i - 1] = i;
sum += (a[i] - a[i - 1]) * (a[i] - a[i - 1]);
}
ans = sum;
for(int x = n; x >= 1; x -= 1){
int i = p[x];
int u = l[i], v = r[i];
if(u != -1){
sum -= (a[i] - a[u]) * (a[i] - a[u]);
r[u] = v;
}
if(v != -1){
sum -= (a[i] - a[v]) * (a[i] - a[v]);
l[v] = u;
}
if(u != -1 && v != -1){
sum += (a[u] - a[v]) * (a[u] - a[v]);
}
ans += sum;
}
cout << ans << "\n";
};
query();
while(q--){
int k;
cin >> k;
k = (k + ans) % n;
rotate(p.begin() + 1, p.begin() + k + 1, p.end());
query();
}
return 0;
}
解法二：魔改树状数组（?）
队友写得魔改树状数组，正着按顺序模拟，qoj 冲过去了。
点击查看代码
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define ll long long
#define PII pair<int,int>
#define PIII pair<PII,PII>
const int N = 2e5+10;
int n,Q;
int a[N];
int cl[N],cr[N];
int lowbit(int x){
return x&(-x);
}
void addl(int x,int v){
for(int i=x;i<=n;i+=lowbit(i)){
cl[i]+=v;
}
}
void addl(int l,int r,int v){
if(r<l)return;
addl(l,v);
addl(r+1,-v);
}
int getl(int x){
int ans=0;
for(int i=x;i>=1;i-=lowbit(i)){
ans+=cl[i];
}
return ans;
}
void addr(int x,int v){
for(int i=x;i<=n;i+=lowbit(i)){
cr[i]+=v;
}
}
void addr(int l,int r,int v){
if(r<l)return;
addr(l,v);
addr(r+1,-v);
}
int getr(int x){
int ans=0;
for(int i=x;i>=1;i-=lowbit(i)){
ans+=cr[i];
}
return ans;
}
deque<int> b;
void solve() {
cin >> n>>Q;
for(int i=1;i<=n;i++){
cin>>a[i];
}
for(int i=1;i<=n;i++){
int x;
cin>>x;
b.push_back(x);
cl[i]=cr[i]=0;
}
ll ans=0,s=0;
for(int i=0;i<n;i++){
int l=getl(b[i]),r=getr(b[i]);
if(!l&&!r){
addr(1,b[i]-1,b[i]);
addl(b[i]+1,n,b[i]);
}
else if(l&&r){
addr(l+1,b[i]-1,-r);
addr(l+1,b[i]-1,b[i]);
addl(b[i]+1,r-1,-l);
addl(b[i]+1,r-1,b[i]);
ans-=(a[r]-a[l])*(a[r]-a[l]);
ans+=(a[r]-a[b[i]])*(a[r]-a[b[i]]);
ans+=(a[l]-a[b[i]])*(a[l]-a[b[i]]);
}else if(l){
addl(b[i]+1,n,-l);
addl(b[i]+1,n,b[i]);
addr(l+1,b[i]-1,b[i]);
ans+=(a[l]-a[b[i]])*(a[l]-a[b[i]]);
}else{
addr(1,b[i]-1,-r);
addr(1,b[i]-1,b[i]);
addl(b[i]+1,r,b[i]);
ans+=(a[r]-a[b[i]])*(a[r]-a[b[i]]);
}
s+=ans;
}
cout<<s<<endl;
a[0]=-1;
for(int o=1;o<=Q;o++){
ll p;
cin>>p;
p+=s;
p%=n;
s=0;
ans=0;
for(int i=0;i<p;i++){
b.push_back(b.front());
b.pop_front();
}
for(int i=1;i<=n;i++){
cl[i]=cr[i]=0;
}
for(int i=0;i<n;i++){
int l=getl(b[i]),r=getr(b[i]);
if(!l&&!r){
addr(1,b[i]-1,b[i]);
addl(b[i]+1,n,b[i]);
}
else if(l&&r){
addr(l+1,b[i]-1,-r);
addr(l+1,b[i]-1,b[i]);
addl(b[i]+1,r-1,-l);
addl(b[i]+1,r-1,b[i]);
ans-=(a[r]-a[l])*(a[r]-a[l]);
ans+=(a[r]-a[b[i]])*(a[r]-a[b[i]]);
ans+=(a[l]-a[b[i]])*(a[l]-a[b[i]]);
}else if(l){
addl(b[i]+1,n,-l);
addl(b[i]+1,n,b[i]);
addr(l+1,b[i]-1,b[i]);
ans+=(a[l]-a[b[i]])*(a[l]-a[b[i]]);
}else{
addr(1,b[i]-1,-r);
addr(1,b[i]-1,b[i]);
addl(b[i]+1,r,b[i]);
ans+=(a[r]-a[b[i]])*(a[r]-a[b[i]]);
}
s+=ans;
}
cout<<s<<endl;
}
}
signed main() {
ios::sync_with_stdio(false), cin.tie(0);
int t = 1;
// cin >> t;
while (t--) {
solve();
}
return 0;
}
J. Let's Play Jigsaw Puzzles!
没读，貌似构造。
点击查看代码
//#pragma GCC optimize(1)
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
//#define int long long
#define endl '\n'
#define ll long long
#define PII pair<int,int>
#define PIII pair<PII,PII>
const int N = 5003;
int m;
int a[N][N];
void solve() {
cin >> m;
if (m == 1) {
int x, y, z, w;
cin >> x >> y >> z >> w;
cout << 1 << endl;
return;
}
vector<PIII>stn; //上左下右
map<int, int>q[3];
for (int i = 1; i <= m * m; i++) {
int x, y, z, w;
cin >> x >> y >> z >> w;
stn.push_back({{x, z}, {y, w}});
if (x != -1)q[0][x]++;
if (z != -1)q[1][z]++;
q[2][x]++;
q[2][y]++;
q[2][z]++;
q[2][w]++;
}
sort(stn.begin(),stn.end());
for (auto [x, y] : q[2]) {
if (y == 2) {
if (q[1][x] == 1&&q[0][x]==1) {
a[1][1] = x;
break;
}
}
}
for (int i = 1; i <= m; i++) {
a[0][i] = a[i][0] = -1;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= m; j++) {
auto[u,v] = *lower_bound(stn.begin(),stn.end(),PIII{{a[i - 1][j], a[i][j - 1]}, {-1, -1}});
auto [x,z]=u;
auto [y,w]=v;
a[i + 1][j] = y;
a[i][j + 1] = w;
}
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= m; j++) {
cout << a[i][j] << " \n"[j == m];
}
}
}
signed main() {
// freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false), cin.tie(0);
int t = 1;
// cin >> t;
while (t--) {
solve();
}
return 0;
}
K. Browser Games
字典树。
字典树上再维护一个每个节点的数量，先将全部字符串加入，后续遍历每个字符串，每新加一个，就从字典树上减去它的计数，然后判断能不能与先前的取的前缀合并掉。
点击查看代码
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
//#define int long long
#define endl '\n'
#define ll long long
#define PII pair<int,int>
#define PIII pair<PII,PII>
const int N = 3e6+10;
int tr[N][28], cnt = 1;
int a[N], vis[N];
int f(char x) {
if (x == '.')return 26;
if (x == '/')return 27;
return x - 'a';
}
vector<int>q;
int ans=0;
void insert(string s, int v) {
int len = s.length(), p = 1;
for (int i = 0; i < len; i++) {
int c = f(s[i]);
if (tr[p][c] == 0)tr[p][c] = ++cnt;
p = tr[p][c];
if(v==-1)q.push_back(p);
a[p] += v;
}
if(v==-1){
if(!a[p]){
vis[p]=1;
ans++;
}
}
if (v == -1) {
for (int i = q.size()-1; i >=0; i--) {
auto p=q[i];
if(!a[p]){
int f=0;
for(int j=0;j<28;j++){
if(tr[p][j]&&vis[tr[p][j]])f=1,vis[tr[p][j]]=0,ans--;
}
if(f){
if(vis[p]){
}else{
vis[p]=1;
ans++;
}
}
}else{
break;
}
}
q.clear();
}
}
int n;
string b[N];
void solve() {
cin >> n;
a[1] = 1;
for (int i = 1; i <= n; i++) {
cin >> b[i];
insert(b[i], 1);
}
for (int i = 1; i <= n; i++) {
insert(b[i], -1);
cout<<ans<<endl;
}
}
signed main() {
// freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false), cin.tie(0);
int t = 1;
// cin >> t;
while (t--) {
solve();
}
return 0;
}
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