AtCoder Beginner Contest 325
A - Takahashi san
#include <bits/stdc++.h>
using namespace std;
#define ll long long
using vi = vector<int>;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s , t;
cin >> s >> t;
cout << s << " san\n";
return 0;
}
B - World Meeting
//#include <bits/stdc++.h>
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
#define int long long
using vi = vector<int>;
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vi w( 24 );
for( int i = 1 , a , b ; i <= n ; i ++ )
cin >> a >> b , w[b] += a;
int res = 0;
for( int l = 0 , r = 8 ; l <= 23 ; l ++ , r ++ ){
int cnt = 0;
for ( int i = l ; i <= r ; i ++ )
cnt += w[ i % 24 ];
res = max( res , cnt );
}
cout << res << "\n";
return 0;
}
C - Sensors
#include <bits/stdc++.h>
using namespace std;
#define ll long long
using vi = vector<int>;
int dx[] = {0,0,1,-1,1,1,-1,-1};
int dy[] = {1,-1,0,0,1,-1,1,-1};
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int H,W;cin >> H >> W;
vector<string> s(H);
for (int i = 0;i < H;i++){
cin >> s[i];
}
vector<vector<int>> vis(H,vector<int>(W));
int res = 0;
auto dd = [&](int x,int y)->void{
queue<pair<int,int>> q;
q.emplace(x,y);
while (!q.empty()){
auto [u,v] = q.front();
q.pop();
if (vis[u][v]) continue;
vis[u][v] = 1;
for (int i = 0;i < 8;i++){
int tx = u + dx[i],ty = v + dy[i];
if (tx < 0 || tx >= H || ty < 0 || ty >= W || vis[tx][ty]) continue;
if (s[tx][ty] == '.') continue;
q.emplace(tx,ty);
}
}
};
for (int i = 0;i < H;i++){
for (int j = 0;j < W;j++){
if (vis[i][j] || s[i][j] == '.') continue;
res++;
dd(i,j);
}
}
cout << res << endl;
return 0;
}
D - Printing Machine
枚举开始时间，然后把所有已经到达的加入到队列中，然后每次贪心的选择结束时间最少的即可。
#include <bits/stdc++.h>
using namespace std;
#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;
const int inf = 1e9, INF = 1e18;
const int mod = 998244353;
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int n;
cin >> n;
vector<pii> p(n);
for (auto &[x, y]: p) cin >> x >> y;
p.emplace_back(LLONG_MAX, LLONG_MAX);
sort(p.begin(), p.end());
multiset<int> cnt;
int now = 0, res = 0;
for (int i = 0; i < n;) {
now = max(now, p[i].first);
for (; i < n and p[i].first <= now; i++)
cnt.insert(p[i].first + p[i].second);
while (!p.empty() and now < p[i].first) {
auto it = cnt.lower_bound(now);
if (it == cnt.end()) break;
cnt.erase(it);
now++, res++;
}
}
cout << res << "\n";
return 0;
}
E - Our clients, please wait a moment
根据汽车火车分别建图，然后求两遍最短路，然后枚举换乘点即可
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
using vi = vector<int>;
const int inf = 1e18;
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n ;
cin >> n;
int a , b , c;
cin >> a >> b >> c;
vector<vi> g( n+1 , vi( n + 1 ) );
auto f = g;
for( int i = 1 ; i <= n ; i ++ ){
for( int j = 1 , x ; j <= n ; j ++ ){
cin >> x;
g[i][j] = x * a;
f[i][j] = x * b + c ;
}
}
vi dis( n+1 , inf );
vi dis1(n + 1,inf);
dis[1] = 0;
priority_queue<pair<int,int> , vector<pair<int,int>> , greater<pair<int,int>>> q;
q.emplace( 0 , 1 );
vi vis( n + 1 );
while( !q.empty() ){
auto [ d , x ] = q.top();
q.pop();
if( vis[x] ) continue;
vis[x] = 1;
for( int y = 1 ; y <= n ; y ++ ){
if( dis[y] <= d + g[x][y] ) continue;
dis[y] = d + g[x][y];
q.emplace( dis[y] , y );
}
}
dis1[n] = 0;
priority_queue<pair<int,int> , vector<pair<int,int>> , greater<pair<int,int>>> q1;
q1.emplace( 0 , n );
vi vis1( n + 1 );
while( !q1.empty() ){
auto [ d , x ] = q1.top();
q1.pop();
if( vis1[x] ) continue;
vis1[x] = 1;
for( int y = 1 ; y <= n ; y ++ ){
if( dis1[y] <= d + f[x][y] ) continue;
dis1[y] = d + f[x][y];
q1.emplace( dis1[y] , y );
}
}
int res = 1e18;
for (int i = 1;i <= n;i++){
res = min(res,dis[i] + dis1[i]);
}
cout << res << endl;
return 0;
}
F - Sensor Optimization Dilemma
\(f[i][j]\)表示前\(i\)个部分用了\(j\)个 1 型监控器最少用多少个 2 型监控器。然后枚举一下\(i\)位用了多少个 1 型监控器即可。
#include <bits/stdc++.h>
using namespace std;
#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;
const int inf = 1e10, INF = 1e18;
const int mod = 998244353;
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int n;
cin >> n;
vi d(n + 1);
for (int i = 1; i <= n; i++) cin >> d[i];
int l1, c1, k1, l2, c2, k2;
cin >> l1 >> c1 >> k1 >> l2 >> c2 >> k2;
vector f(n + 1, vi(k1 + 1, 0));
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= k1; j++) {
f[i][j] = inf;
for (int k = 0, t; k <= j; k++) {
t = d[i] - k * l1, t = (t + l2 - 1) / l2;
f[i][j] = min(f[i][j], f[i - 1][j - k] + t);
if (t == 0) break;
}
}
}
int res = INF;
for (int i = 0; i <= k1; i++)
if (f[n][i] <= k2) res = min(res, c1 * i + c2 * f[n][i]);
if( res == INF ) cout << "-1\n";
else cout << res << "\n";
return 0;
}
G - offence
一个不太好想的区间 dp
\(f[l][r]\)表示\([l,r]\)经过若干次操作后所能剩下的最短长度，则答案即为\(f[0][n-1]\)
转移首先考虑用两个子区间拼起来的最优解，其次再考虑端点为o或f的情况。
#include <bits/stdc++.h>
using namespace std;
#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;
const int inf = 1e10, INF = 1e18;
const int mod = 998244353;
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int K;
string s;
cin >> s >> K;
int n = s.size();
vector f(n + 1, vi(n + 1));
for (int l = 0; l < n; l++)
for (int r = 0; r < n; r++)
f[l][r] = r - l + 1;
for (int len = 2; len <= n; len++) {
for (int l = 0, r = len - 1; r < n; l++, r++) {
if (len == 2) {
if (s[l] == 'o' and s[r] == 'f') f[l][r] = 0;
} else {
for (int k = l; k < r; k++)
f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r]);
if (s[r] == 'f') {
for (int k = l; k < r; k++)
if (s[k] == 'o' and f[k + 1][r - 1] == 0)
f[l][r] = min(f[l][r], (k == 0) ? 0 : f[l][k - 1]);
}
if (s[l] == 'o') {
for (int k = l + 1; k <= r; k++) {
if (s[k] == 'f' and f[l + 1][k - 1] == 0)
f[l][r] = min(f[l][r], max(f[k + 1][r] - K, 0ll));
}
}
}
}
}
cout << f[0][n - 1] << "\n";
return 0;
}
