Contest 2050 and Codeforces Round #718 (Div. 1 + Div. 2）abcd

 

2050 cf比赛 传送门

 

A   Sum of 2050  standard input/output  1 s, 256 MB Submit Add to favourites     x16113

//可以直接暴力数组枚举 也可以 一步步算出来

B   Morning Joggingstandard input/output 1 s, 256 MB    Submit Add to favourites     x8733

//其实就是n*m 的矩阵， 需要把每一列的最小值相加最小， 做法是把所有的数排个序， 然后按从小到大 放在相应的列上， 其他数就随便放 用一个多元组 tuple存每个数的大小和行列位置， 然后只按大小排序  接着前m个数放在不同的列上（t神真强

C   Fillomino 2 standard input/output 1 s, 256 MB   Submit Add to favourites     x7570

//开始想半天不会， 然后看了t神做法， 直接把每个数 一直往左边放 放不下了就往下放  然后再往左边放 可以证明这个是有效的

D   Explorer Space standard input/output 2 s, 256 MB    Submit Add to favourites     x4122

直接暴力搜索每个点的最小k/2路径和， dp记录下（ O(nmk)

#include <bits/stdc++.h>

using namespace std;

#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
#define all(v) (v).begin(), (v).end()
#define nl "\n"
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define sz(v) ((int)(v).size())
//#define LOCAL

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;

#ifdef LOCAL
//#include "pretty_print.h"
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__)
#else
#define dbg(...) 42
#endif

template <typename T> T sqr(T x) { return x * x; }
template <typename T> T abs(T x) { return x < 0? -x : x; }
template <typename T> T gcd(T a, T b) { return b? gcd(b, a % b) : a; }
template <typename T> bool chmin(T &x, const T& y) { if (x > y) { x = y; return true; } return false; }
template <typename T> bool chmax(T &x, const T& y) { if (x < y) { x = y; return true; } return false; }

auto random_address = [] { char *p = new char; delete p; return (uint64_t) p; };
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count() * (random_address() | 1));
mt19937_64 rngll(chrono::steady_clock::now().time_since_epoch().count() * (random_address() | 1));

 

void taskA() {
    int t; cin >> t;
    while(t--) {
        ll n; cin >> n;
        if(n%2050 != 0) {
            cout << "-1\n"; 
            continue;
        }
        n /= 2050;
        int ans = 0;
        while(n != 0) {
            ans += (n%10);
            n /= 10;
        }
        cout << ans << "\n";
    }
    return; 
}
void taskA1() {
    int t; cin >> t;
    while(t--) {
        ll a[] = {
                  2050, 
                  20500, 
                  205000, 
                  2050000, 
                  20500000, 
                  205000000, 
                  2050000000, 
                  20500000000,
                  205000000000, 
                  2050000000000,    
                  20500000000000, 
                  205000000000000,
                  2050000000000000,
                  20500000000000000,
                  205000000000000000
              };
        ll n; cin >> n;
        int f = 0;
        ll ans = 0;
        while(1) {
            if(n == 0) break;
            else if(n < a[0]) {
                cout << "-1\n";
                f = 1;
                break;
            }
            else {
                for(int i = 14; i >= 0; i--)
                    if(n >= a[i]) 
                        ans += (n/a[i]), n = n%a[i];
            }
        }
        if(!f)
            cout << ans << "\n";
    }
    return;
}

 

void taskB() {
    int t; cin >> t;
    while(t--) {
        int h, w; cin >> h >> w;
         vector<vector<int>> a(h, vector<int>(w));
         vector<vector<int>> b(h, vector<int>(w, -1));
         
         vector<tuple<int, int, int> > tem;
         _for(i,0,h)  _for(j,0,w) 
            cin >> a[i][j], tem.emplace_back(a[i][j], i, j);
        sort(all(tem));
        _for(j,0,w) {
            int i = get<1>(tem[j]);
            b[i][j] = get<0>(tem[j]);
        }
        _for(x,w,h*w) {
            int i = get<1>(tem[x]);
            _for(j,0,w)
            if(b[i][j] == -1) {
                b[i][j] = get<0>(tem[x]);
                break;
            }
        }

        _for(i,0,h) {
            _for(j,0,w) {
                if(j != 0) cout << " ";
                cout << b[i][j];    
            } cout << "\n";
        }
    }
    return;
}

 

void taskC() {
    int n; cin >> n;
    vector<vector<int>> a(n, vector<int>(n, 0));
    _for(i,0,n) cin >> a[i][i];

    _for(i,0,n) {
        int tmp = a[i][i]-1;
        int x = i, y = i; 
        while(tmp--) {
            
            if(y>0 and a[x][y-1] == 0) y--;
            else x++;
            a[x][y] = a[i][i];
        }
    }

    _for(i,0,n) {
        _for(j,0,i+1) {
            if(j != 0) cout << " ";
            cout << a[i][j];
        } cout << "\n";
    }
    return ;
}

 

void taskD() {
    int n,m,k; cin >> n >> m >> k;
    vector<vector<int> > row(n, vector<int> (m-1, 0));
    vector<vector<int> > col(n-1, vector<int> (m, 0));
 
    _for(i,0,n) _for(j,0,m-1) cin >> row[i][j];
 
    _for(i,0,n-1) _for(j,0,m) cin >> col[i][j];
 
    vector<vector<int> > ans(n, vector<int>(m, 0));
    if(k%2 == 1) {
        ans.assign(n, vector<int>(m, -1));
    } else {
        int N = (int)(1e9);
        vector<vector<int> > new_ans(n, vector<int>(m, 0));
        _for(l,0,k/2) {
            _for(i,0,n) 
                _for(j,0,m) {
                    new_ans[i][j] = N;
                    if(i > 0)   new_ans[i][j] = min(ans[i-1][j]+2*col[i-1][j], new_ans[i][j]);//Down
 
                    if(j > 0)   new_ans[i][j] = min(ans[i][j-1]+2*row[i][j-1], new_ans[i][j]);//Right
 
                    if(i < n-1) new_ans[i][j] = min(ans[i+1][j]+2*col[i][j], new_ans[i][j]);//Up
 
                    if(j < m-1) new_ans[i][j] = min(ans[i][j+1]+2*row[i][j], new_ans[i][j]);//Left
                }
            swap(ans, new_ans);
        }
        
    }
 
    _for(i,0,n)
    {
        _for(j,0,m) 
        {
            if(j > 0) cout << " ";
            cout << ans[i][j];
        } cout << "\n";
    }
 
    return;
}

 

int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);

    taskD();
    return 0;
}