2024黑龙江CCPC

2024黑龙江省赛

B. String

• 思路:栈模拟
• 代码：

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-28 23:32:45
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;

string s;
signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> s;
	stack<char>stk;
	for(auto it : s){
		stk.push(it);
		if(stk.size() >= 3){
			char a = stk.top(); stk.pop();
			char b = stk.top(); stk.pop();
			char c = stk.top(); stk.pop();
			if(!(a == b && b == c)){
				stk.push(c);stk.push(b);stk.push(a);
			}
		}
	}
	string ans;
	while(stk.size()){
		ans += stk.top();
		stk.pop();
	}
	reverse(all(ans));
	if(!ans.size()) cout << "NAN";
	else cout << ans;
	return 0;
}

D. Card Game

• 思路：贪心，把A从大到小排序，把B不为-1的从小到大排序，按顺序出牌，这种方法最优
• 代码：

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-28 23:42:10
*/
#include <bits/stdc++.h>
// #define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;

const int N = 100010;
int a[N], b[N], ah, bh, n;

void solve(){
	cin >> n >> ah >> bh;
	for(int i = 1; i <= n; i ++ ) cin >> a[i];
	vector<int>bb;
	int cntb = 0;
	for(int i = 1; i <= n; i ++ ){
		cin >> b[i];
		if(b[i] != -1) bb.pb(b[i]);
		else cntb ++;
	} 
	sort(all(bb));
	while(cntb -- ){
		bb.pb(-1);
	}
	sort(a + 1, a + 1 + n, greater<int>());
	for(int i = 1; i <= n; i ++ ){
		if(bb[i - 1] >= 0 && a[i] != -1) ah -= bb[i - 1];
		if(a[i] >= 0 && bb[i - 1] != -1) bh -= a[i];
		if(ah > 0 && bh <= 0){
			cout << "yes\n";
			return;
		}
		if(ah <= 0){
			cout << "no\n";
			return;
		}
	}
	cout << "no\n";
}

signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
	int tt = 1;
	cin >> tt;
	while(tt -- ){
		solve();
	}
	return 0;
}

F. Photography

• 思路 ：先预处理每一个点最大的三个邻接点，存下标。然后暴力枚举每两条边，如果能形成一条链，就从链的两端进行拓展出最大的那个点，这是处理长度大于3的情况。还要注意考虑长度为1, 2的情况。这题有孤点。

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-29 03:30:40
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;
 
const int N = 10010;
int a[N], big[N][3];
int n, m, ans, now;
vector<PII>q[N];
vector<PII>edges;	

int work(int same1, int same2, int diff1, int diff2){
	int four = -1;
	int now = a[same1] + a[diff1] + a[diff2];
	bool f = 1;
	if(big[diff1][0] == same1 || big[diff1][0] == diff2 || big[diff1][0] == -1){
		if(big[diff1][1] == same1 || big[diff1][1] == diff2 || big[diff1][1] == -1){
			if(big[diff1][2] == -1 || big[diff1][2] == diff2 || big[diff1][2] == same1){
				f = 0;
			}else now += a[big[diff1][2]], four = big[diff1][2];
		}else now += a[big[diff1][1]], four = big[diff1][1];
	}else now += a[big[diff1][0]], four = big[diff1][0];
	if(big[diff2][0] == same1 || big[diff2][0] == diff1 || big[diff2][0] == four || big[diff2][0] == -1){
		if(big[diff2][1] == same1 || big[diff2][1] == diff1 || big[diff2][1] == four || big[diff2][1] == -1){
			if(big[diff2][2] == same1 || big[diff2][2] == diff1 || big[diff2][2] == four || big[diff2][2] == -1){
				f = 0;
			}else now += a[big[diff2][2]];
		}else now += a[big[diff2][1]];
	}else now += a[big[diff2][0]];
	return now;
}

signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n >> m;
	for(int i = 1; i <= n; i ++ ){
		cin >> a[i];
		for(int j = 0; j <= 2; j ++ ) big[i][j] = -1;
	}
	for(int i = 1; i <= m; i ++ ){
		int x, y; cin >> x >> y;
		q[x].pb({a[y], y});
		q[y].pb({a[x], x});
		edges.pb({x, y});
		if(x != y) ans = max(ans, a[x] + a[y]);
	}
	for(int i = 1; i <= n; i ++ ) ans = max(ans, a[i]);
	for(int i = 1; i <= n; i ++ ){
		sort(all(q[i]));
		int len = q[i].size();
		if(len >= 1) big[i][0] = q[i][len - 1].se;
		if(len >= 2) big[i][1] = q[i][len - 2].se;
		if(len >= 3) big[i][2] = q[i][len - 3].se;
	}
	for(int i = 0; i < m; i ++ ){
		for(int j = 0; j < m; j ++ ){
			if(i == j) continue;
			int x = edges[i].fi, y = edges[i].se, xx = edges[j].fi, yy = edges[j].se;
			if(x == xx) ans = max(ans, work(x, xx, y, yy));
			if(x == y) ans = max(ans, work(x, y, xx, yy));
			if(y == xx) ans = max(ans, work(y, xx, x, yy));
			if(y == yy) ans = max(ans, work(y, yy, x, xx));
		}
	}
	cout << ans << endl;
	return 0;
}

I. This is an easy problem

• 思路：模拟
• 代码

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-28 23:26:33
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;

int n;

signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n;
	int cnt = 0;
	while(n){
		cnt += n % 2;
		n /= 2;
	}
	cout << cnt;
	return 0;
}

J. Trade

• 思路：dp, $dp[i][j]$表示，走到第$i,j$的路径上的$\sum b[i][j]$，转移方式类似摘花生那题并要结合条件判断当前点是否是利润非负。最后看最后一行或者最后一列有没有小于无穷大的数即可。
• 代码

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-29 00:46:32
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;

const int N = 1010;
LL a[N][N], b[N][N], dp[N][N];
int n, m;

signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n >> m;
	for(int i = 1; i <= n; i ++ ){
		for(int j = 1; j <= m; j ++ ){
			cin >> a[i][j];
			dp[i][j] = 9e18;
		}
	}
	for(int i = 1; i <= n; i ++ ){
		for(int j = 1; j <= m; j ++ ){
			cin >> b[i][j];
		}
	}
	dp[1][1] = b[1][1];
	for(int i = 1; i <= n; i ++ ){
		for(int j = 1; j <= n; j ++ ){
			if(i == 1 && j == 1) continue;
			if(i >= 2) dp[i][j] = min(dp[i - 1][j], dp[i][j]);
			if(j >= 2) dp[i][j] = min(dp[i][j - 1], dp[i][j]);
			dp[i][j] += b[i][j];
			if(a[i][j] - a[1][1] - dp[i][j] < 0) dp[i][j] = 9e18;
		}
	}
	bool f = 0;
	for(int i = 1; i <= n; i ++ ){
		if(dp[i][m] < 9e18) f = 1;
	}
	for(int i = 1; i <= m; i ++ ){
		if(dp[n][i] < 9e18) f = 1;
	}
	if(f) cout << "YES";
	else cout << "NO";
	return 0;
}

K. Puzzle

• 思路：纯模拟
• 代码：

#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;

int a[10];
set<int>s;
signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> a[1] >> a[2] >> a[3] >> a[4];
	sort(a + 1, a + 5);
	set<int>s;
	do{
		for(int i = 1; i <= 3; i ++ ){
		for(int j = 1; j <= 3; j ++ ){
			for(int k = 1; k <= 3; k ++ ){
				if(i == 1 && j == 1 && k == 1) s.insert(a[1] + a[2] +a[3] + a[4]);
				if(i == 1 && j == 1 && k == 2) s.insert(a[1] + a[2] +a[3] - a[4]);
				if(i == 1 && j == 1 && k == 3) s.insert(a[1] + a[2] +a[3] * a[4]);
				if(i == 1 && j == 2 && k == 1) s.insert(a[1] + a[2] -a[3] + a[4]);
				if(i == 1 && j == 2 && k == 2) s.insert(a[1] + a[2] -a[3] - a[4]);
				if(i == 1 && j == 2 && k == 3) s.insert(a[1] + a[2] -a[3] * a[4]);
				if(i == 1 && j == 3 && k == 1) s.insert(a[1] + a[2] *a[3] + a[4]);
				if(i == 1 && j == 3 && k == 2) s.insert(a[1] + a[2] *a[3] - a[4]);
				if(i == 1 && j == 3 && k == 3) s.insert(a[1] + a[2] *a[3] * a[4]);

				if(i == 2 && j == 1 && k == 1) s.insert(a[1] - a[2] +a[3] + a[4]);
				if(i == 2 && j == 1 && k == 2) s.insert(a[1] - a[2] +a[3] - a[4]);
				if(i == 2 && j == 1 && k == 3) s.insert(a[1] - a[2] +a[3] * a[4]);
				if(i == 2 && j == 2 && k == 1) s.insert(a[1] - a[2] -a[3] + a[4]);
				if(i == 2 && j == 2 && k == 2) s.insert(a[1] - a[2] -a[3] - a[4]);
				if(i == 2 && j == 2 && k == 3) s.insert(a[1] - a[2] -a[3] * a[4]);
				if(i == 2 && j == 3 && k == 1) s.insert(a[1] - a[2] *a[3] + a[4]);
				if(i == 2 && j == 3 && k == 2) s.insert(a[1] - a[2] *a[3] - a[4]);
				if(i == 2 && j == 3 && k == 3) s.insert(a[1] - a[2] *a[3] * a[4]);

				if(i == 3 && j == 1 && k == 1) s.insert(a[1] * a[2] +a[3] + a[4]);
				if(i == 3 && j == 1 && k == 2) s.insert(a[1] * a[2] +a[3] - a[4]);
				if(i == 3 && j == 1 && k == 3) s.insert(a[1] * a[2] +a[3] * a[4]);
				if(i == 3 && j == 2 && k == 1) s.insert(a[1] * a[2] -a[3] + a[4]);
				if(i == 3 && j == 2 && k == 2) s.insert(a[1] * a[2] -a[3] - a[4]);
				if(i == 3 && j == 2 && k == 3) s.insert(a[1] * a[2] -a[3] * a[4]);
				if(i == 3 && j == 3 && k == 1) s.insert(a[1] * a[2] *a[3] + a[4]);
				if(i == 3 && j == 3 && k == 2) s.insert(a[1] * a[2] *a[3] - a[4]);
				if(i == 3 && j == 3 && k == 3) s.insert(a[1] * a[2] *a[3] * a[4]);
			}
		}
	}
	}while(next_permutation(a + 1,a + 1 + 4));
	cout << sz(s);
	return 0;
}