2017 BJ ICPC 石子合并变种 向量基本功及分类考察

E

模拟

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
int cat[105];
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > que;
int main()
{
        int m, n, x;
        while (cin >> m >> n >> x)
        {
                int anser1 = m;
                int anser2 = 0;
                for (int i = 1; i <= n; i++)
                {
                        scanf("%d", &cat[i]);
                        que.push(make_pair(0, cat[i]));
                }
                while (!que.empty())
                {
                        pair<int, int> cnt = que.top();
                        if (cnt.first >= x)
                        {
                                que.pop();
                                continue;
                        }
                        else
                        {
                                que.pop();
                                if (anser1 > 0)
                                {
                                        cnt.first += cnt.second;
                                        if (cnt.first <= x)
                                        {
                                                anser1--;
                                        }
                                        else
                                        {
                                                anser1--;
                                                anser2++;
                                        }
                                        que.push(cnt);
                                }
                        }
                }
                cout << anser1 << " " << anser2 << endl;
        }
}

F

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{0, 1}, { 1, 0}, { 0, -1}, { -1, 0}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
char f[105][105];
char ans[100005];
char anser[105][105];
int pop = 0;
int dx, dy;
int dir;
int n;
bool check(int x, int y)
{
        if (x > n || x < 1)
        {
                return false;
        }
        if (y > n || y < 1)
        {
                return false;
        }
        if (anser[x][y] != '.')
        {
                return false;
        }
        return true;
}
int main()
{
        while (cin >> n)
        {
                for (int i = 1; i <= 100; i++)
                {
                        for (int j = 1; j <= 100; j++)
                        {
                                anser[i][j] = '.';
                        }
                }
                dir = 0;
                dx = dy = 1;
                pop = 0;
                for (int i = 1; i <= n; i++)
                {
                        scanf("%s", f[i] + 1);
                }
                for (int i = 2; i <= n + 1; i++)
                {
                        if (i % 2 == 0)
                        {
                                for (int j = i - 1; j >= 1; j--)
                                {
                                        ans[++pop] = f[j][i - j];
                                }
                        }
                        else
                        {
                                for (int j = 1; j <= i - 1; j++)
                                {
                                        ans[++pop] = f[j][i - j];
                                }

                        }
                }
                for (int i = n + 2; i <= 2 * n; i++)
                {
                        if (i % 2 == 0)
                        {
                                for (int j = n; j >= i - n; j--)
                                {
                                        ans[++pop] = f[j][i - j];
                                }

                        }
                        else
                        {
                                for (int j = i - n; j <= n; j++)
                                {
                                        ans[++pop] = f[j][i - j];
                                }
                        }
                }
//                for (int i = 1; i <= pop; i++)
//                {
//                        cout << ans[i];
//                }
//                cout << endl;
                for (int i = 1; i <= pop; i++)
                {
                        //cout<<dx<<" "<<dy<<endl;
                        anser[dx][dy] = ans[i];
                        if (!check(dx + turn[dir][0], dy + turn[dir][1]))
                        {
                                dir = (dir + 1) % 4;
                        }
                        dx = dx + turn[dir][0];
                        dy = dy + turn[dir][1];
                }
                for (int i = 1; i <= n; i++)
                {
                        for (int j = 1; j <= n; j++)
                        {
                                cout << anser[i][j];
                        }
                        cout << endl;
                }
        }
}

J

石子合并 不过合并的范围由2变为L-R（2<=L<=R<=N） 问你最少的花费是多少