Codeforces Round #791 (Div. 2)解题报告
A. AvtoBus
题意：有两种车，一种车4轮，另一种6轮，给你轮子的数量，猜出能恰好组成的车辆数量的最大和最小，如果凑不出就输出-1
分析: 显而易见奇数个轮子肯定不恰好能凑成若干车辆，再考虑偶数情况，将6看作两个2加一个2，4看作两个2，那么所有大于4的偶数都可以看作是若干个2组成的，那么所有的偶数都可以由4和6组成
故输出-1的条件为x < 4 || x & 1,最大的计算：肯定能组成的话，先尽量组成4轮的，如果有剩余，肯定剩2，随便找一个4轮的安成6轮的，所以maxv = x / 4,最小的计算：先尽量组成6轮的，如果有剩余那么剩下的不是2就是4，拉出一辆6轮的组成俩4轮或一4一6，车辆数 + 1，所以minv = x / 6 + (x % 6 != 0);
ac 代码
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include <sstream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl '\n'
#define pi 3.14159265358979323846
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 200010,M = 500010,INF = 0x3f3f3f3f,mod = 1e9 + 7 ;
const double INFF = 0x7f7f7f7f7f7f7f7f;
int n,m,x,y,k,idx;
int row[N],col[N];
set<int> r;
set<int> co;
int main()
{
ios;
int t;
cin >> t;
while(t --)
{
LL x;
cin >> x;
if(x < 4 || x & 1) cout << -1 << endl;
else
{
LL maxv = x / 4, minv = x / 6 + (x % 6 != 0);
cout << minv << " " << maxv << endl;
}
}
return 0;
}
B. Stone Age Problem
题意：给你一个数组，定义以下两个操作：
1.令a[pos] = x;
2.令所有数变成x
每次操作都要输出当前数组的总和
分析：模拟
ac代码
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include <sstream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl '\n'
#define pi 3.14159265358979323846
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2000010,M = 500010,INF = 0x3f3f3f3f,mod = 1e9 + 7 ;
const double INFF = 0x7f7f7f7f7f7f7f7f;
int n,k;
int a[N];
string s,p;
LL sum = 0,last = 0;
bool flag = false,st[N];
int main()
{
ios;
cin >> n >> k;
for(int i = 1;i <= n;i ++) cin >> a[i],sum += a[i];
while(k --)
{
int t,pos,x;
cin >> t;
if(t == 1)
{
cin >> pos >> x;
if(!flag)
{
sum -= a[pos];
a[pos] = x;
sum += a[pos];
}
else
{
if(!st[pos])
{
sum -= last - x;
a[pos] = x;
st[pos] = true;
}
else
{
sum -= a[pos] - x;
a[pos] = x;
}
}
cout << sum << endl;
}
else
{
flag = true;
memset(st,0,sizeof(bool) * (n + 4));
cin >> x;
sum = 1LL * x * n;
last = x;
cout << sum << endl;
}
}
return 0;
}
C. Rooks Defenders
题意：给定一个\(n * n\) 的棋盘，定义以下三个操作：
1.在（x,y）处放一个棋子
2.在（x,y）处取走棋子
3.问 以\((x_1,y_1)\)为左上角\((x_2,y_2)\)为右下角的区域是否每个格子都被攻击
攻击条件:一个格子所在行或所在列有棋子就被攻击
在进行操作3后，需回答
分析：维护空行和空列两个有序集合，
version 1 二叉搜索树
采用二叉搜索树来维护，每次询问对有序行集合进行查找大于x1的最小元素，对有序列集合进行查找大于y1的最小元素，如果行大于x2，或列大于y2，说明整个区域都被攻击，
ac代码
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include <sstream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl '\n'
#define pi 3.14159265358979323846
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 200010,M = 500010,INF = 0x3f3f3f3f,mod = 1e9 + 7 ;
const double INFF = 0x7f7f7f7f7f7f7f7f;
int n,m,x,y,k,idx;
int row[N],col[N];
set<int> r;
set<int> co;
int main()
{
ios;
cin >> n >> k;
for(int i = 1;i <= n;i ++) co.insert(i),r.insert(i);
while(k --)
{
int t,a,b,c,d;
cin >> t;
if(t == 1)
{
cin >> a >> b;
row[a] ++,col[b] ++;
if(row[a] == 1) r.erase(a);
if(col[b] == 1) co.erase(b);
}
else if(t == 2)
{
cin >> a >> b;
row[a] -- , col[b] --;
if(!row[a]) r.insert(a);
if(!col[b]) co.insert(b);
}
else
{
cin >> a >> b >> c >> d;
set<int>::iterator A = r.lower_bound(a),B = co.lower_bound(b);
if(A == r.end() || B == co.end()) cout << "YES" << endl;
else if(* A > c || * B > d) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
return 0;
}
version 2 树状数组
用两个树状数组维护行与列的非空行
ac代码
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include <sstream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl '\n'
#define pi 3.14159265358979323846
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 200010,M = 500010,INF = 0x3f3f3f3f,mod = 1e9 + 7 ;
const double INFF = 0x7f7f7f7f7f7f7f7f;
int n,m,x,y,k,idx;
int row[N],col[N];
int tr1[N],tr2[N];
int lowbit(int x)
{
return x & -x;
}
void add(int tr[],int x,int v)
{
for(int i = x;i < N;i += lowbit(i)) tr[i] += v;
}
int query(int tr[],int x)
{
int res = 0;
for(int i = x;i;i -= lowbit(i)) res += tr[i];
return res;
}
int main()
{
ios;
cin >> n >> k;
while(k --)
{
int t,a,b,c,d;
cin >> t;
if(t == 1)
{
cin >> a >> b;
row[a] ++, col[b] ++;
if(row[a] == 1) add(tr1,a,1);
if(col[b] == 1) add(tr2,b,1);
}
else if(t == 2)
{
cin >> a >> b;
row[a] --,col[b] --;
if(row[a] == 0) add(tr1,a,-1);
if(col[b] == 0) add(tr2,b,-1);
}
else
{
cin >> a >> b >> c >> d;
x = query(tr1,c) - query(tr1,a - 1);
y = query(tr2,d) - query(tr2,b - 1);
if(x >= c - a + 1 || y >= d - b + 1) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
return 0;
}
D. Toss a Coin to Your Graph...
题意：给定一个有向图，每个节点上有点权，问所有长度大于等于k - 1的路径中点权最大值的最小值
分析：
version 1 二分 + 记忆化搜索
答案无非就是点权中的某一个，所以另开空间再存一遍点权排个序，进行二分。找长度大于等于k - 1的路径的最小最大值
ac代码
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include <sstream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl '\n'
#define pb(x) push_back(x)
#define pi 3.14159265358979323846
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 200010,M = 500010,INF = 0x3f3f3f3f,mod = 1e9 + 7 ;
const double INFF = 0x7f7f7f7f7f7f7f7f;
int n,m,w[N],a[N];
LL k;
int h[N],e[N],ne[N],idx;
LL f[N];
bool st[N];
void add(int a,int b)
{
e[idx] = b,ne[idx] = h[a],h[a] = idx ++;
}
LL dfs(int u,int c)
{
if(f[u])
{
if(f[u] == -1) return 0;
else return f[u];
}
if(w[u] > c)
{
f[u] = -1;
return 0;
}
LL ans = 0;
st[u] = true;
for(int i = h[u];i != -1;i = ne[i])
{
int j = e[i];
if(st[j]) ans = k;
else ans = max(ans,dfs(j,c));
if(ans >= k) break;
}
st[u] = false;
return f[u] = ans + 1;
}
bool check(int x)
{
memset(st,0,sizeof (bool) * (n + 1));
memset(f,0,sizeof (LL) * (n + 1));
LL ans = 0;
for(int i = 1;i <= n;i ++)
{
ans = max(ans,dfs(i,x));
if(ans >= k) break;
}
return ans >= k;
}
int main()
{
ios;
cin >> n >> m >> k;
memset(h,-1,sizeof (int) * (n + 1));
for(int i = 1;i <= n;i ++) cin >> a[i],w[i] = a[i];
while(m --)
{
int c,b;
cin >> c >> b;
add(c,b);
}
sort(a + 1,a + 1 + n);
int l = 1,r = n;
while(l < r)
{
int mid = l + r >> 1;
if(check(a[mid])) r = mid;
else l = mid + 1;
}
if(check(a[l])) cout << a[l] << endl;
else cout << -1 << endl;
return 0;
}
version 2 拓扑排序 + 二分
拓扑排序判环，顺便找最长链，每次check跑一遍topsort
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include <sstream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#define x first
#define y second
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl '\n'
#define pb push_back
#define pi 3.14159265358979323846
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 200010,M = 500010,INF = 0x3f3f3f3f,mod = 1e9 + 7 ;
const double INFF = 0x7f7f7f7f7f7f7f7f;
int n,m,w[N],a[N],d[N],q[N];
LL k;
int h[N],e[N],ne[N],idx;
LL f[N];
bool st[N];
vector<PII> edges;
void add(int a,int b)
{
e[idx] = b,ne[idx] = h[a],h[a] = idx ++;
}
bool topsort()
{
int hh = 0, tt = -1;
fill(f + 1,f + 1 + n, 1);
for(int i = 1;i <= n;i ++) if(!d[i]) q[++ tt] = i;
while(hh <= tt)
{
int t = q[hh ++];
for(int i = h[t];i != -1;i = ne[i])
{
int j = e[i];
f[j] = max(f[j],f[t] + 1);
if( -- d[j] == 0) q[++ tt] = j;
}
}
return tt < n - 1 || * max_element(f + 1,f + 1 + n) >= k;
}
bool check(int x)
{
idx = 0;
memset(h,-1,sizeof (int) * (n + 1));
memset(d,0,sizeof (int) * (n + 1));
for(auto i:edges)
{
if(w[i.x] <= x && w[i.y] <= x) add(i.x,i.y),d[i.y] ++;
}
return topsort();
}
int main()
{
ios;
cin >> n >> m >> k;
memset(h,-1,sizeof (int) * (n + 1));
for(int i = 1;i <= n;i ++) cin >> a[i],w[i] = a[i];
while(m --)
{
int a,b;
cin >> a >> b;
edges.pb({a,b});
}
sort(a + 1,a + 1 + n);
int l = 1,r = n;
while(l < r)
{
int mid = l + r >> 1;
if(check(a[mid])) r = mid;
else l = mid + 1;
}
if(check(a[l]))cout << a[l] << endl;
else cout << -1 << endl;
return 0;
}
