Polaris.AI Programming Contest 2025（AtCoder Beginner Contest 429）

D - On AtCoder Conference

环形 二分

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f#define int long longint n, m, c;
int a[500010];
int s[2*500010];
map<int, int> mp;void solve(){cin >> n >> m >> c;vector<int> vec;vec.pb(0); vec.pb(m);for(int i = 1; i <= n; i ++){cin >> a[i];vec.pb(a[i]); mp[a[i]] ++;}sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()),vec.end());int nn = vec.size();for(int i = 0; i < nn; i ++) vec.pb(vec[i]);for(int i = 0; i < nn + nn; i ++){if(i == 0) s[i] = 0;else s[i] = s[i - 1] + mp[vec[i]];}int ans = 0;for(int i = 0; i < nn - 1; i ++){int l = i + 1; int r = nn + nn - 1;while(l < r){int mid = (l + r) >> 1;if(s[mid] - s[i] >= c) r = mid;else l = mid + 1;}// cout << (s[l] - s[i]) << ' ' << vec[i + 1] - 1 - vec[i] + 1 << '\n';ans += (s[l] - s[i]) * (vec[i + 1] - 1 - vec[i] + 1);}cout << ans << '\n';
}signed main(){std::ios::sync_with_stdio(false);int T = 1; //cin >> T;while(T--){solve();}
}

E - Hit and Away

对于每一个危险点，输出从一个安全点经过他到达另一个安全点的最小距离
这种跨着的，类似树的直径，随便四个更新都可以

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f#define int long longconst int N = 200010;
vector<int> G[N];
int vis[N];
int d1[N], p1[N], d2[N], p2[N];void solve(){int n, m; cin >> n >> m;while (m --){int u, v; cin >> u >> v;G[u].pb(v); G[v].pb(u);}string s; cin >> s; s = "a" + s;queue<int> q;for(int i = 1; i <= n; i ++){d1[i] = INF; p1[i] = -1;d2[i] = INF; p2[i] = -1;}for(int i = 1; i <= n; i ++){if(s[i] == 'S'){d1[i] = 0; p1[i] = i;q.push(i);}}while(!q.empty()){int u = q.front(); q.pop();for(auto v:G[u]){if(d1[v] > d1[u] + 1 && p1[v] != p1[u]){d2[v] = d1[v];p2[v] = p1[v];d1[v] = d1[u] + 1;p1[v] = p1[u];q.push(v);}if(d2[v] > d1[u] + 1 && p1[v] != p1[u]){d2[v] = d1[u] + 1;p2[v] = p1[u];q.push(v);}if(d1[v] > d2[u] + 1 && p1[v] != p2[u]){d2[v] = d1[v];p2[v] = p1[v];d1[v] = d2[u] + 1;p1[v] = p2[u];q.push(v);}if(d2[v] > d2[u] + 1 && p1[v] != p2[u]){d2[v] = d2[u] + 1;p2[v] = p2[u];q.push(v);}}}for(int i = 1; i <= n; i ++){//   cout << d1[i] << ' ' << p1[i] << ' ' << d2[i] << ' ' << p2[i] << '\n';if(s[i] == 'D'){cout << d1[i] + d2[i] << '\n';}}
}signed main(){std::ios::sync_with_stdio(false);int T = 1; //cin >> T;while(T--){solve();}
}

F - Shortest Path Query

三行，所以不可能往回走
分治，用线段树每次单点修改

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f#define int long longconst int N = 200010;
char a[4][N];struct Tree{int l, r;int d[4][4];
}tr[4*N];void emp(Tree& u){for(int i = 1; i <= 3; i ++)for(int j = 1; j <= 3; j ++) u.d[i][j] = INF;
}
void emp(int u){emp(tr[u]);
}
void pushup(Tree& u, Tree & l, Tree & r){ emp(u);u.l = l.l; u.r = r.r;for(int i = 1; i <= 3; i ++){for(int j = 1; j <= 3; j ++){for(int k = 1; k <= 3; k ++){if(a[k][l.r] == '.'){u.d[i][j] = min(u.d[i][j], l.d[i][k] + 1 + r.d[k][j]);}}}}
}
void pushup(int u){pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}void chg(int u){emp(u);for(int i = 1; i <= 3; i ++){int j = i;while(j >= 1 && a[j][tr[u].l] == '.'){tr[u].d[i][j] = abs(i - j);j --;} j = i;while(j <= 3 && a[j][tr[u].l] == '.') {tr[u].d[i][j] = abs(i - j);j ++;}}
}void bui(int u, int l, int r){emp(u);tr[u].l = l; tr[u].r = r;if(l == r){chg(u);return ;}int mid = (l + r) >> 1;bui(u*2, l, mid);bui(u*2 + 1, mid + 1, r);pushup(u);
}void modify(int u, int x, int y){if(tr[u].l == tr[u].r && tr[u].l == y){ chg(u);return ;}int mid = (tr[u].l + tr[u].r) >> 1;if(y <= mid) modify(u*2, x, y);else modify(u * 2 + 1, x, y);pushup(u);
}void solve(){int n; cin >> n;for(int i = 1; i <= 3; i ++){string s; cin >> s;for(int j = 1; j <= n; j ++) a[i][j] = s[j - 1];}bui(1, 1, n);int q; cin >> q;while(q --){int x, y; cin >> x >> y;if(a[x][y] == '.')a[x][y] = '#';else a[x][y] = '.';modify(1, x, y);if(tr[1].d[1][3] > INF / 2) cout << -1 << '\n';else cout << tr[1].d[1][3] << '\n';}}signed main(){std::ios::sync_with_stdio(false);int T = 1;// cin >> T;while(T--){solve();}
}