2020-2021 ICPC Southwestern European Regional Contest (SWERC 2020)
A. Gratitude
字符串再加上个pair排序即可
#include <bits/stdc++.h>
using namespace std;
#define PII pair<int,int>
#define fi first
#define se second
#define pb push_back
#define ll long long
const int N=1e6+10;
int n,k;
map<string,int> mp,pos;
vector<pair<string,pair<int,int>>> v;
string s;
int main(){
cin>>n>>k;
getchar();
for(int i=1;i<=3*n;++i){
getline(cin,s);
mp[s]++;
pos[s]=i;
}
int cnt=0;
for(auto w:mp){
v.pb({w.fi,{w.se,pos[w.fi]}});
cnt++;
}
sort(v.begin(),v.end(),[&](pair<string,pair<int,int>> a,pair<string,pair<int,int>> b){
if(a.se.fi!=b.se.fi) return a.se.fi>b.se.fi;
return a.se.se>b.se.se;
});
for(int i=0;i<min(k,cnt);++i){
cout<<v[i].fi<<endl;
}
return 0;
}
C. Safe Distance
队友写的。。
#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef pair<double, double> PDD;
const double eps = 1e-6;
const int N = 1e3+10;
double X, Y;
PDD q[N];
int n;
int p[N];
double ans;
int find(int x){
if(p[x]!=x) return p[x]=find(p[x]);
return p[x];
}
double get_dist(PDD a, PDD b){
double dx = a.x - b.x;
double dy = a.y - b.y;
return sqrt(dx * dx + dy * dy);
}
int check(double r){
int v, w;
for(int i = 1 ;i <= n + 4 ;i++) p[i]=i;
for(int i = 1; i <= n; i ++){
if(q[i].x<r){//zuo
v=find(i); w=find(n+4);
if(v!=w) p[v]=w;
}
if(X - q[i].x < r){//you
v=find(i), w=find(n+2);
if(v!=w) p[v]=w;
}
if(q[i].y < r){//xia
v=find(i), w=find(n+3);
if(v!=w) p[v]=w;
}
if(Y - q[i].y < r){//shang
v=find(i), w=find(n+1);
if(v!=w) p[v]=w;
}
for(int j = i + 1; j <= n ; j++){
double dis = get_dist(q[i], q[j]);
if(dis < r * 2){
v=find(i), w=find(j);
if(v!=w) p[w]=v;
}
}
}
if(find(n+1)!=find(n+2) && find(n+1)!=find(n+3) && find(n+3)!=find(n+4) && find(n+2)!=find(n+4)) return 1;
/* if(find(n+1) == find(n+2)) return -1;
else if(find(n+1) == find(n+3)) return -1;
else if(find(n+3) == find(n+4)) return -1;
else if(find(n+2) == find(n+4)) return -1;
else return 1;*/
return -1;
}
int main()
{
scanf("%lf %lf",&X, &Y);
scanf("%d",&n);
for(int i = 1; i <= n;i ++){
scanf("%lf %lf",&q[i].x, &q[i].y);
}
double l = 0, r = max(X, Y);
while(r - l >= eps){
double mid = (l + r) / 2;
// printf("mid=%lf\n",mid);
if(check(mid)==-1){
r=mid;
}else{
l=mid;
}
}
printf("%lf\n",l);
return 0;
}
D. Jogging
队友过的，好像就是个最短路板子题
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#include<iomanip>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define eps 1.0e18
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define rep(x,y,z) for(int x = y; x <= z; x++)
#define per(x,y,z) for(int x = y; x >= z; x--)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll qpow(ll a,ll b,ll mod){ll res = 1;while(b){if(b&1){res=res*a%mod;}b>>=1;a=a*a%mod;}return res%mod;}
const ll mod = 1e9+7;
const ll maxn = 1e5+5;
vector<pair<int,int>> vec[200000];
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
map<pair<int,int>,int> mp;
bool vis[200000];
int dis[200000];
void dijkstra(int n)
{
for(int i = 1; i < n; i++){
dis[i] = 1000000007;
vis[i] = false;
}
dis[0] = 0;
vis[0] = false;
pq.push({0,0});
while(!pq.empty()){
pair<int,int> p = pq.top();
pq.pop();
int id = p.second;
int d = p.first;
if(vis[id])continue;
vis[id] = true;
for(pair<int,int> u:vec[id]){
if(dis[u.first]>dis[id]+u.second){
dis[u.first]=dis[id]+u.second;
pq.push({dis[u.first],u.first});
}
}
}
}
int main()
{
ios::sync_with_stdio(false);
int n,m,l,r;
cin>>n>>m>>l>>r;
for(int i = 1,u,v,x; i <= m; i++){
cin>>u>>v>>x;
vec[u].push_back({v,x});
vec[v].push_back({u,x});
}
dijkstra(n);
ll ans = 0;
for(int i = 0; i < n; i++){
if(dis[i]*2>=r)continue;
int len = vec[i].size();
for(int j = 0; j < len; j++){
int a = min(i,vec[i][j].first);
int b = max(i,vec[i][j].first);
if(mp[{a,b}])continue;
ans++;
mp[{a,b}]++;
}
}
cout<<ans<<endl;
return 0;
}
E. Cakes
签到
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#include<iomanip>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define eps 1.0e18
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define rep(x,y,z) for(int x = y; x <= z; x++)
#define per(x,y,z) for(int x = y; x >= z; x--)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll qpow(ll a,ll b,ll mod){ll res = 1;while(b){if(b&1){res=res*a%mod;}b>>=1;a=a*a%mod;}return res%mod;}
const ll mod = 1e9+7;
const ll maxn = 1e5+5;
int main()
{
ios::sync_with_stdio(false);
int n;
cin>>n;
int ans = 1000000000;
for(int i = 1,x,y; i <= n; i++){
cin>>x>>y;
ans = min(ans,y/x);
}
cout<<ans<<endl;
return 0;
}
H. Figurines
-
题意：每天可以在一排横着的货架上添加或删去物品，定义：\(x_{i+1}=x_i+y_i\),\(x_0=0\),\(y_i\)表示第\(i\)天货架上不小于\(x_i\)的物品数，给你\(n\)个询问，第\(i\)个询问问你第\(d_i\)天的\(x_i\)值。
-
题解：\(x_i\)是线性得到的，我们唯一需要求的就是第\(d_i\)天时，货架上不小于\(x_i\)的物品数，显而易见，对历史区间查询，主席树即可啦～，注意这题的读入。
-
代码：
#include <bits/stdc++.h> #define ll long long #define fi first #define se second #define pb push_back #define me memset #define rep(a,b,c) for(int a=b;a<=c;++a) #define per(a,b,c) for(int a=b;a>=c;--a) const int N = 5e5 + 10; const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; using namespace std; typedef pair<int,int> PII; typedef pair<ll,ll> PLL; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll lcm(ll a,ll b) {return a/gcd(a,b)*b;} int n; int root[N],idx; int mp[N]; struct Node{ int l,r; int cnt; }tr[N<<4]; int build(int l,int r){ int p=++idx; if(l==r) return p; int mid=(l+r)>>1; tr[p].l=build(l,mid); tr[p].r=build(mid+1,r); return p; } int update(int p,int l,int r,int x,int k){ int q=++idx; tr[q]=tr[p]; tr[q].cnt+=k; if(l==r) return q; int mid=(l+r)>>1; if(x<=mid) tr[q].l=update(tr[p].l,l,mid,x,k); else tr[q].r=update(tr[p].r,mid+1,r,x,k); return q; } int query(int u,int l,int r,int x){ if(l>=x) return tr[u].cnt; if(l==r) return tr[u].cnt; int mid=(l+r)>>1; int sum=0; if(x<=mid) sum+=query(tr[u].l,l,mid,x); sum+=query(tr[u].r,mid+1,r,x); return sum; } int main() { cin>>n; root[0]=build(1,n); getchar(); int cnt=0; for(int i=1;i<=n;++i){ string s; getline(cin,s); int num=0; int flag=1; for(int j=0;j<(int)s.size();++j){ if(s[j]==' '){ cnt++; root[cnt]=update(root[cnt-1],1,n,num,flag); num=0; continue; } if(j==(int)s.size()-1){ cnt++; num=num*10+s[j]-'0'; root[cnt]=update(root[cnt-1],1,n,num,flag); break; } if(s[j]=='+'){ flag=1; } else if(s[j]=='-')flag=-1; else num=num*10+s[j]-'0'; } mp[i]=cnt; } int x=0; for(int i=1;i<=n;++i){ int d; cin>>d; int res=query(root[mp[d]],1,n,x); x=(x+res)%n; } cout<<x<<endl; return 0; }
I. Emails
-
题意：一张图，每个点都可以储存其他的点的信息，每次，每个点会把自己和其他相邻点的信息传给所有相邻的点，问你最少操作多少次使得每个点都储存了其他所有点的信息,允许有+1的误差。
-
题解：首先手玩几个样例，对于两个点\((u,v)\)不难发现一定在\(\lceil log_2(u->v)\rceil\)次后相互得到对方信息,那么对于整张图的最坏情况为\(\lceil log_2(diam)\rceil\)(直径).现在假设我们任意选一个点出发，其最远路径为\(D\)，那么有：
\(1.D\le diam\le 2*D\).
\(2.\log(D)\le \log(diam)\le \log(2*D)+1\)
\(3.log(diam)\le\log(2*D)+1\le\log(diam)+1\)
结合1、3得到答案为：\(\lceil \log(2*D)\rceil +1\).
-
代码：
#include <bits/stdc++.h> #define ll long long #define fi first #define se second #define pb push_back #define me memset #define rep(a,b,c) for(int a=b;a<=c;++a) #define per(a,b,c) for(int a=b;a>=c;--a) const int N = 3e6 + 10; const int mod = 1e9 + 7; const int INF = 0x3f3f3f3f; using namespace std; typedef pair<int,int> PII; typedef pair<ll,ll> PLL; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll lcm(ll a,ll b) {return a/gcd(a,b)*b;} int n,m; vector<int> edge[N]; bool vis[N]; int dis[N]; int main() { scanf("%d %d",&n,&m); me(dis,INF,sizeof(dis)); for(int i=1;i<=m;++i){ int u,v; scanf("%d %d",&u,&v); edge[u].pb(v); edge[v].pb(u); } int mx=0; queue<int> q; dis[1]=0; q.push(1); while(!q.empty()){ int u=q.front(); q.pop(); if(vis[u]) continue; vis[u]=true; for(auto to:edge[u]){ if(dis[to]>dis[u]+1){ dis[to]=dis[u]+1; q.push(to); mx=max(mx,dis[to]); } } } for(int i=1;i<=n;++i){ if(dis[i]==INF){ puts("-1"); return 0; } } printf("%d\n",(int)ceil(log2(mx))+1); return 0; }
K. Unique Activities
二分答案，然后用hash来存字串判断即可
#include <bits/stdc++.h>
using namespace std;
#define PII pair<int,int>
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long
const int N=1e6+10;
int n;
string s;
ull base=196613;
ull h[N],p[N];
void init(){
p[0]=1;
for(int i=1;i<=n;++i){
h[i]=h[i-1]*base+s[i]-'A';
p[i]=p[i-1]*base;
}
}
bool check(int x){
unordered_map<ull,int> mp;
for(int i=1;i<=n;++i){
if(i>=x){
mp[h[i]-h[i-x]*p[x]]++;
}
}
for(auto w:mp){
if(w.se==1){
return true;
}
}
return false;
}
int main(){
cin>>s;
n=(int)s.size();
s="`"+s;
init();
int l=1,r=n;
while(l<r){
int mid=(l+r)>>1;
if(check(mid)) r=mid;
else l=mid+1;
}
unordered_map<ull,int> mp;
for(int i=1;i<=n;++i){
if(i>=r){
mp[h[i]-h[i-r]*p[r]]++;
}
}
int ans=1;
for(int i=1;i<=n;++i){
if(i>=r && mp[h[i]-h[i-r]*p[r]]==1){
ans=i;
break;
}
}
for(int i=ans-r+1;i<=ans;++i) cout<<s[i];
return 0;
}
L. Restaurants
待补