AtCoder Beginner Contest 245
A - Good morning
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
int a , b , c , d;
int ta , ao;
cin >> a >> b >> c >> d;
ta = a * 60 * 60 + b * 60;
ao = c * 60 * 60 + d * 60 + 1 ;
if( ta < ao ) printf("Takahashi\n");
else printf("Aoki\n");
return 0;
}
B - Mex
#include <bits/stdc++.h>
using namespace std;
int read(){
int x = 0 , f = 1 , ch = getchar();
while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
if( ch == '-' ) f = -1 , ch = getchar();
while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
return x * f;
}
int32_t main() {
int n = read();
vector<int> a(2005);
for( int x ; n ; n -- ) x = read() , a[x] ++;
for( int i = 0 ; i <= 2001 ; i ++ )
if( a[i] == 0 ) return cout << i , 0;
return 0;
}
C - Choose Elements
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read(), k = read();
vector<int> a(n), b(n);
for (int &i: a) i = read();
for (int &i: b) i = read();
vector<array<int, 2> > f(n);
f[0][1] = f[0][0] = 1;
for (int i = 1; i < n; i++) {
if( f[i-1][1] + f[i-1][0] == 0 ) break;
if (f[i - 1][0]) {
if (abs(a[i] - a[i - 1]) <= k) f[i][0] = 1;
if (abs(b[i] - a[i - 1]) <= k) f[i][1] = 1;
}
if (f[i - 1][1]) {
if (abs(a[i] - b[i - 1]) <= k) f[i][0] = 1;
if (abs(b[i] - b[i - 1]) <= k) f[i][1] = 1;
}
}
if( f[n-1][0] || f[n-1][1] ) cout << "Yes\n";
else cout << "No\n";
return 0;
}
D - Polynomial division
多项式除法，列一个数式，手模一下就有了
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define mp make_pair
int32_t main() {
int n = read(), m = read();
vector<int> a(n + 1), b(m + 1), c(n + m + 1);
for (auto &i: a) i = read();
for (auto &i: c) i = read();
for( int i = m ; i >= 0 ; i -- ){
b[i] = c[i+n] / a[n];
for( int j = 0 ; j < n ; j ++ )
c[i+j] -= b[i] * a[j];
}
for( int i = 0 ; i <= m ; i ++ )
printf("%d " , b[i] );
return 0;
}
E - Wrapping Chocolate
把盒子和巧克力豆按照宽度从大到小排序，然后开始枚举巧克力，首先把所有宽度大于当前巧克力的盒子的长度都加入到set中，然后把大于等当前巧克力长度的最小盒子删掉，如果删不掉就输出No
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read(), m = read();
vector<pair<int, int>> a(n), b(m);
for (int i = 0; i < n; i++) a[i].first = read();
for (int i = 0; i < n; i++) a[i].second = read();
for (int i = 0; i < m; i++) b[i].first = read();
for (int i = 0; i < m; i++) b[i].second = read();
sort(a.begin(), a.end(), greater<pair<int, int>>());
sort(b.begin(), b.end(), greater<pair<int, int>>());
multiset<int> st;
for( int i = 0 , j = 0 ; i < n ; i ++ ){
while( j < m && a[i].first <= b[j].first ) st.insert(b[j].second) , j ++;
if( st.empty() ) return cout << "No" , 0;
auto it = st.lower_bound(a[i].second);
if( it == st.end() ) return cout << "No" , 0;
st.erase(it);
}
cout << "Yes";
return 0;
}
F - Endless Walk
实际上就是问有多少个点可以走到环上，首先出度为 0 的点，一定走不到环上，其次是只能走到这些出度为 0 的点也是不行的。这个过程和拓扑排序很接近，所以做法就是反向建边，然后跑一遍拓扑排序，没有删掉的点就是答案。
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read(), m = read() , res = n;
vector<vector<int> > e(n + 1);
vector<int> deg(n+1);
for( int u , v ; m ; m -- ){
u = read() , v = read() , deg[u] ++ , e[v].push_back(u);
}
queue<int> q;
for( int i = 1 ; i <= n ; i ++ )
if( deg[i] == 0 ) q.push(i);
while( !q.empty() ){
auto u = q.front(); q.pop();
res --;
for( auto v : e[u] ){
deg[v] --;
if( deg[v] == 0 ) q.push(v);
}
}
cout << res << "\n";
return 0;
}
G - Foreign Friends
其实最简单的想法肯定是按照颜色分类，然后跑最短路，然后统计答案。
答案最短路次短路还是太难了，抄了抄 dls 的做法。
做法就是按照二进制位吧颜色进行分类，然后跑多远最短路，更新二进制位不同的点的答案。按照二进制位分类的话最多只有 30 多种情况，很快就跑完了。
#include<bits/stdc++.h>
using namespace std;
#define int long long
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read(), m = read(), k = read(), l = read();
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) a[i] = read();
vector<int> b(l + 1);
for (int i = 1; i <= l; i++) b[i] = read();
vector<vector<pair<int, int> > > e(n + 1);
vector<int> res(n + 1, LLONG_MAX);
for (int u, v, w; m; m--) {
u = read(), v = read(), w = read();
e[u].emplace_back(v, w), e[v].emplace_back(u, w);
}
for (int bit = 0; bit <= ceil(log2(k)); bit++) {
for (int t = 0; t < 2; t++) {
vector<int> dis(n + 1, LLONG_MAX);
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
vector<bool> vis(n+1 , false);
for (auto i: b)
if (((a[i] >> bit) & 1) == t)
dis[i] = 0, q.emplace(0, i);
while (!q.empty()) {
auto [d, u] = q.top();
q.pop();
if( vis[u] ) continue;
vis[u] = true;
for( auto [ v , w ] : e[u] ){
if( dis[v] > d + w ) dis[v] = d + w , q.emplace( dis[v] , v );
}
}
for( int i = 1 ; i <= n ; i ++ ){
if( ((a[i]>>bit)&1) == 1-t)
res[i] = min( res[i] , dis[i] );
}
}
}
for( int i = 1 ; i <= n ; i ++ )
printf("%lld " , (res[i] == LLONG_MAX ? -1 : res[i]) );
return 0;
}
