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登录—专业IT笔试面试备考平台_牛客网

注意: 本题有负数, 所以虽然是求最大值, 但是要初始化数组为负无穷, 同时初始化状态的起点为 0

#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<PII, int> PIII;
const int inf = 0x3f3f3f3f;
const ll infinf = 0x3f3f3f3f3f3f3f3f;

const int N = 1010;
ll w[N][N], f[N][N], finv[N][N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m;
    cin >> n >> m;

    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++) 
            cin >> w[i][j];
        
    // 这题有负数, 所以要初始化, 初始化成负无穷
    memset(f,-0x3f,sizeof f);
    memset(finv,-0x3f,sizeof finv);
    // 初始化完要初始化状态的起点
    f[0][1] = 0;
    f[1][0] = 0;
    finv[n + 1][m] = 0;
    finv[n][m + 1] = 0;
    
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            f[i][j] = max(f[i - 1][j], f[i][j - 1]) + w[i][j];
    
    for (int i = n; i >= 1; i --) 
        for (int j = m; j >= 1; j --) 
            finv[i][j] = max(finv[i + 1][j], finv[i][j + 1]) + w[i][j];


    int t;
    cin >> t;

    while (t --) {
        int k;
        cin >> k;

        int x, y;
        vector<PII> a;
        for (int i = 1; i <= k; i ++) {
            cin >> x >> y;
            a.push_back({x, y});
        }
        // 因为点数只有5个, 所以暴力枚举所有传送可能
        ll maxn = -1e18;
        for (int i = 0; i < a.size(); i ++) {
            for (int j = 0; j < a.size(); j ++) {
                if (i != j) {
                    maxn = max(maxn, f[a[i].first][a[i].second] + finv[a[j].first][a[j].second]);
                }
            }
        }

        cout << max(maxn, f[n][m]) << endl;
        
    }

    return 0;
}