第34次CCF计算机软件能力认证

(第一题)矩阵重塑（其一）

#include<iostream>
#include<algorithm>

using namespace std;

int n, m, p, q;

int main()
{
    cin >> n >> m >> p >> q;
    int k = 0;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
        {
            int x;
            cin >> x;
            printf("%d ", x);
            if(++k == q){
                k = 0;
                printf("\n");
            }
        }
}

(第二题)矩阵重塑（其二）

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 1e4 + 10;

int n, m, t;
int w[N];
int temp[N];

int main()
{
    cin >> n >> m >> t;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            cin >> w[i * m + j];

    while(t--)
    {
        int op, a, b;
        cin >> op >> a >> b;
        if(op == 1)
        {
            n = a, m = b;
        }
        else if(op == 2)
        {
            for(int i = 0; i < n; i++)
                for(int j = 0; j < m; j++)
                    temp[j * n + i] = w[i * m + j];
            memcpy(w, temp, sizeof temp);
            swap(n, m);
        }
        else if(op == 3)
        {
            cout << w[a * m + b] << endl;
        }

    }
}

(第四题)货物调度

参考题解：2024-06 第34次 ccf-csp T4 货物调度 100pts 题解_第34次ccfcsp答案-CSDN博客

#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;

typedef long long ll;

const int N = 1e3 + 10, M = 4e4 + 10;

int n, m, v;
vector<int> w[N];//w[i][j]:第i个仓库中,第j大的货物价值
int b[N], c[N];//仓库的基本费用与计件费用
ll dp[N][M];//dp[i][j]:在前i个仓库中取货物,在总费用为j的情况下,获得的最大总价值,M为计算的最大总花费

bool cmp(int a, int b)
{
    return a > b;
}

int main()
{
    cin >> n >> m >> v;
//    for(int i=0;i<M;i++)dp[0][i]=0;
    for(int i = 1; i <= n; i++) cin >> b[i] >> c[i];
    for(int i = 1; i <= m; i++)
    {
        int a, b;
        cin >> a >> b;
        b++;
        w[b].push_back(a);
    }

    for(int i = 1; i <= n; i++) sort(w[i].begin(), w[i].end(), cmp);//按货物价值降序排序,通过贪心思路进行DP获得最优解

    int res = 0x3f3f3f3f;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 0; j < M; j++)
        {
            dp[i][j] = dp[i - 1][j];
            ll sum = 0;
            for(int k = 1; k <= w[i].size(); k++)
            {
                if(j < b[i] + k * c[i]) break;
                sum += w[i][k - 1];
                dp[i][j] = max(dp[i][j], dp[i - 1][j - (b[i] + k * c[i])] + sum - (b[i] + k * c[i]));
            }
            if(dp[i][j] >= v) res = min(res, j);
        }
    }

    cout << res;
}