Educational Codeforces Round 27
A. Chess Tourney
拿第一队最弱的和第二队最强的比就好了
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 2e5+7;
void solve(){
int n; sci(n);
vi A(2*n); for(int &x : A) sci(x);
sort(all(A));
if(A[n]>A[n-1]) cout << "YES" << endl;
else cout << "NO" << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
B.Luba And The Ticket
每次拿变化可以最大的数字去变
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 2e5+7;
char s[10];
void solve(){
cin >> s;
int s1 = 0, s2 = 0;
for(int i = 0; i < 3; i++) s1 += s[i] - '0', s2 += s[i+3] - '0';
int ret = 0;
while(s1!=s2){
if(s1<s2){
int a = 0, b = 3;
if(s[1]<s[0]) a = 1;
if(s[2]<s[a]) a = 2;
if(s[4]>s[3]) b = 4;
if(s[5]>s[b]) b = 5;
if('9'-s[a] > s[b]-'0'){
if(s2-s1<='9'-s[a]){
ret++;
break;
}else{
s1 += '9' - s[a];
s[a] = '9';
ret++;
}
}else{
if(s2-s1<=s[b]-'0'){
ret++;
break;
}else{
s2 -= s[b] - '0';
s[b] = '0';
ret++;
}
}
}else{
swap(s1,s2);
for(int i = 0; i < 3; i++) swap(s[i],s[i+3]);
}
}
cout << ret << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
C. Two TVs
维护两个指针，分别表示当前看完的结束时间，把所有区间按左端点排序，遇到看不了的电视就输出\(NO\)
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 2e5+7;
int n;
void solve(){
sci(n);
vector<pii> A(n);
for(auto &p : A) sci(p.first), sci(p.second);
sort(all(A));
int a = -1, b = -1;
for(auto p : A){
if(p.first > a) a = p.second;
else if(p.first > b) b = p.second;
else{
cout << "NO" << endl;
return;
}
}
cout << "YES" << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
D. Driving Test
用栈存限速，记一下过了多少个的不能超车，加速了之后从栈顶把超速的都删掉，加上删掉的数量
超车了就加上全部的记下的标记
遇到可超车就把标记全部删掉
模拟即可
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 2e5+7;
void solve(){
int n; sci(n);
int cnt = 0;
int noovertake = 0;
stack<int> speedlimit;
int carspeed = 0;
while(n--){
int tp; sci(tp);
if(tp==1){
sci(carspeed);
while(!speedlimit.empty() and speedlimit.top()<carspeed) cnt++, speedlimit.pop();
}else if(tp==2){
cnt += noovertake;
noovertake = 0;
}else if(tp==3){
int x; sci(x);
if(carspeed>x) cnt++;
else speedlimit.push(x);
}else if(tp==4) noovertake = 0;
else if(tp==5) while(!speedlimit.empty()) speedlimit.pop();
else noovertake++;
}
cout << cnt << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
E. Fire in the City
二分时间
关于\(check\)，可以用离散化然后扫描线来做，具体就是找到没有被覆盖的点的\(x\)最大最小值和\(y\)的最大最小值
然后判断能否用一个边长为\(mid*2+1\)的正方形覆盖四个边界点
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 5e4+7;
int n, m, k;
vector<pii> vec;
struct SegTree{
int cover[MAXN<<2], l[MAXN<<2], r[MAXN<<2];
bool ept[MAXN<<2];
#define ls(rt) rt << 1
#define rs(rt) rt << 1 | 1
void pushup(int rt){
if(cover[rt]) ept[rt] = false;
else if(l[rt] + 1 == r[rt]) ept[rt] = (cover[rt]==0);
else ept[rt] = (ept[ls(rt)] or ept[rs(rt)]);
}
void build(int L, int R, int rt = 1){
l[rt] = L; r[rt] = R;
ept[rt] = true; cover[rt] = 0;
if(l[rt] + 1 == r[rt]) return;
int mid = (L + R) >> 1;
build(L,mid,ls(rt)); build(mid,R,rs(rt));
}
void modify(int L, int R, int x, int rt = 1){
if(L>=r[rt] or l[rt]>=R) return;
if(L<=l[rt] and r[rt]<=R){
cover[rt] += x;
pushup(rt);
return;
}
modify(L,R,x,ls(rt)); modify(L,R,x,rs(rt));
pushup(rt);
}
bool checkempty(){ return ept[1]; }
}ST;
vector<pair<pii,int> > seg[MAXN];
bool check(int mid){
vector<pair<pii,pii> > rect;
for(auto &p : vec){
int &x = p.first, &y = p.second;
rect << make_pair(make_pair(max(1,x-mid),max(1,y-mid)),make_pair(min(n,x+mid),min(m,y+mid)));
}
vector<int> vecx, vecy;
for(auto &p : rect){
vecx << p.first.first << p.second.first;
vecy << p.first.second << p.second.second;
if(p.first.first!=1) vecx << (p.first.first-1);
if(p.first.second!=1) vecy << (p.first.second-1);
if(p.second.first!=n) vecx << (p.second.first+1);
if(p.second.second!=m) vecy << (p.second.second+1);
}
sort(all(vecx)); sort(all(vecy));
vecx.erase(unique(all(vecx)),vecx.end());
vecy.erase(unique(all(vecy)),vecy.end());
for(auto &p : rect){
p.first.first = lower_bound(all(vecx),p.first.first) - vecx.begin() + 1;
p.second.first = lower_bound(all(vecx),p.second.first) - vecx.begin() + 1;
p.first.second = lower_bound(all(vecy),p.first.second) - vecy.begin() + 1;
p.second.second = lower_bound(all(vecy),p.second.second) - vecy.begin() + 1;
}
int x1 = n + 1, y1 = m + 1, x2 = 0, y2 = 0;
if(vecx[0]!=1 or vecy.front()!=1 or vecy.back()!=m) x1 = 0;
else{
for(int i = 1; i <= (int)vecx.size(); i++) seg[i].clear();
for(auto &p : rect){
seg[p.first.first] << make_pair(make_pair(p.first.second,p.second.second),1);
seg[p.second.first+1] << make_pair(make_pair(p.first.second,p.second.second),-1);
}
ST.build(1,vecy.size() + 1);
for(int i = 1; i <= (int)vecx.size(); i++){
for(auto &p : seg[i]) ST.modify(p.first.first,p.first.second+1,p.second);
if(ST.checkempty()){
x1 = vecx[i-1] - 1;
break;
}
}
}
if(vecx.back()!=n or vecy.front()!=1 or vecy.back()!=m) x2 = n + 1;
else{
for(int i = 1; i <= (int)vecx.size(); i++) seg[i].clear();
for(auto &p : rect){
seg[p.first.first-1] << make_pair(make_pair(p.first.second,p.second.second),-1);
seg[p.second.first] << make_pair(make_pair(p.first.second,p.second.second),1);
}
ST.build(1,vecy.size() + 1);
for(int i = (int)vecx.size(); i >= 1; i--){
for(auto &p : seg[i]) ST.modify(p.first.first,p.first.second+1,p.second);
if(ST.checkempty()){
x2 = vecx[i-1] + 1;
break;
}
}
}
if(vecy[0]!=1 or vecx.front()!=1 or vecx.back()!=n) y1 = 0;
else{
for(int i = 1; i <= (int)vecy.size(); i++) seg[i].clear();
for(auto &p : rect){
seg[p.first.second] << make_pair(make_pair(p.first.first,p.second.first),1);
seg[p.second.second+1] << make_pair(make_pair(p.first.first,p.second.first),-1);
}
ST.build(1,vecx.size() + 1);
for(int i = 1; i <= (int)vecy.size(); i++){
for(auto &p : seg[i]) ST.modify(p.first.first,p.first.second+1,p.second);
if(ST.checkempty()){
y1 = vecy[i-1] - 1;
break;
}
}
}
if(vecy.back()!=m or vecx.front()!=1 or vecx.back()!=n) y2 = m+1;
else{
for(int i = 1; i <= (int)vecy.size(); i++) seg[i].clear();
for(auto &p : rect){
seg[p.first.second-1] << make_pair(make_pair(p.first.first,p.second.first),-1);
seg[p.second.second] << make_pair(make_pair(p.first.first,p.second.first),1);
}
ST.build(1,vecx.size() + 1);
for(int i = (int)vecy.size(); i >= 1; i--){
for(auto &p : seg[i]) ST.modify(p.first.first,p.first.second+1,p.second);
if(ST.checkempty()){
y2 = vecy[i-1] + 1;
break;
}
}
}
int dx = x2 - x1 - 1, dy = y2 - y1 - 1;
return max(dx/2,dy/2) <= mid;
}
void solve(){
sci(n); sci(m); sci(k);
vec.resize(k);
for(auto &p : vec) sci(p.first), sci(p.second);
int l = 0, r = max(n,m);
while(l<=r){
int mid = (l + r) >> 1;
if(check(mid)) r = mid - 1;
else l = mid + 1;
}
cout << l << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
F. Guards In The Storehouse
显然是要状压\(dp\)的
\(dp[i][j][mask][x][y]\)
\(i\)表示第\(i\)行，\(j\)表示第\(j\)列，\(mask\)表示到当前位置每一列上是否有没有被墙堵住的守卫，\(x\)表示同一行左边是否有没有被墙堵住的守卫，\(y\)表示是否有没有被守卫保护到的区域
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 1<<15;
const int MOD = 1e9+7;
const int N = 15;
int n, m, f[2][MAXN][2][2];
char buf[1<<8][1<<8], s[1<<8][1<<8];
void solve(){
sci(n); sci(m);
for(int i = 0; i < n; i++) scanf("%s",buf[i]);
if(n>=m) for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) s[i][j] = buf[i][j];
else{
for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) s[j][i] = buf[i][j];
swap(n,m);
}
int tag = 0;
f[0][0][0][0] = 1;
auto setzero = [&](int x, int y){ return (x>>y&1) ? (x ^ (1 << y)) : x; };
for(int i = 0; i < n; i++) for(int j = 0; j < m; j++){
tag ^= 1;
memset(f[tag],0,sizeof(f[tag]));
for(int msk = 0; msk < (1 << m); msk++){
for(int x = 0; x < 2; x++) for(int y = 0; y < 2; y++){
if(s[i][j]=='x') f[tag][setzero(msk,j)][0][y] = (f[tag][setzero(msk,j)][0][y] + f[tag^1][msk][x][y]) % MOD;
else{
if(j==m-1){
if(x or (msk>>j&1)) f[tag][msk][0][y] = (f[tag][msk][0][y] + f[tag^1][msk][x][y]) % MOD; // no guard in this cell
f[tag][msk|(1<<j)][0][y] = (f[tag][msk|(1<<j)][0][y] + f[tag^1][msk][x][y]) % MOD; // set a guard
if(!x and !(msk>>j&1) and !y) f[tag][msk][0][1] = (f[tag][msk][0][1] + f[tag^1][msk][x][y]) % MOD;
}else{
if(x or (msk>>j&1)) f[tag][msk][x][y] = (f[tag][msk][x][y] + f[tag^1][msk][x][y]) % MOD; // no guard in this cell
f[tag][msk|(1<<j)][1][y] = (f[tag][msk|(1<<j)][1][y] + f[tag^1][msk][x][y]) % MOD; // set a guard
if(!x and !(msk>>j&1) and !y) f[tag][msk][0][1] = (f[tag][msk][0][1] + f[tag^1][msk][x][y]) % MOD;
}
}
}
}
}
int ret = 0;
for(int msk = 0; msk < (1 << m); msk++) for(int x = 0; x < 2; x++) for(int y = 0; y < 2; y++) ret = (ret + f[tag][msk][x][y]) % MOD;
cout << ret << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
G. Shortest Path Problem?
可以发现如果存在环的话，这个环上的异或和是可以在任何时候被选上使用的，可以考虑从\(1\)到环上一点，然后绕一圈在回去就好了，所以可以把所有环上的值放到一个线性基里面，然后我们找到任意一条\(1\)到\(n\)的路径，把这个路径异或和放到线性基里面取个\(\min\)就好了
考虑如何把所有环的异或和算出来，可以在建\(dfs\)树的过程中算，遇到一条非树边\(u,v\)之后，我们把\(dis_u\oplus dis_v\oplus w_{uv}\)放到线性基里面
view code
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define endl "\n"
#define LL long long int
#define vi vector<int>
#define vl vector<LL>
#define all(V) V.begin(),V.end()
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%I64d",&x)
#define pii pair<int,int>
#define pll pair<LL,LL>
#ifndef ONLINE_JUDGE
#define cout cerr
#endif
#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))
#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))
#define debug(x) cerr << #x << " = " << x << endl
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }
template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }
const int MAXN = 2e5+7;
int n, m, vis[MAXN], dis[MAXN];
vector<pii> G[MAXN];
int base[30];
void ins(int x){
for(int i = 29; ~i; i--){
if(x>>i&1){
if(!base[i]){
base[i] = x;
break;
}
x ^= base[i];
}
}
}
int get_min(int x){
for(int i = 29; ~i; i--){
if(x>>i&1 and base[i]) x ^= base[i];
}
return x;
}
void dfs(int u, int par){
vis[u] = true;
bool findpar = false;
for(auto &e : G[u]){
int v = e.first, w = e.second;
if(v==par and !findpar){
findpar = true;
continue;
}
if(!vis[v]){
dis[v] = dis[u] ^ w;
dfs(v,u);
}else ins(dis[u] ^ dis[v] ^ w);
}
}
void solve(){
sci(n); sci(m);
for(int i = 1; i <= m; i++){
int u, v, w; sci(u); sci(v); sci(w);
G[u] << pii(v,w); G[v] << pii(u,w);
}
dfs(1,0);
cout << get_min(dis[n]) << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
