CF_2055_C. The Trail

题目链接：Problem - C - Codeforces

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题目大意：

给一个 n 行 m 列的矩形网格 e(−1e6≤e[i][j]​≤1e6) ，

给一个长度为 n+m−2 的字符串 s ( 只含D 和 R )，从左上角到右下角的路径

路径上的数字都为 0 你需要将整个路径上的 0 填上数字，使得每一行，每一列的和都相同。

$输出：任意方案填补后的矩阵$

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思路：

直接设 $n$ 行 $m$ 列的矩阵每行之和或每列之和为 $0$

$然后根据路径走一遍，路径上的填补的数\; =\; 这行或这列之和的相反数$

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#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 1008, mod = 998244353;

int n, m;
int e[N][N];
int s1[N], s2[N];

void solve() {

    mst(s1, 0), mst(s2, 0);
    string s;
    cin >> n >> m >> s;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> e[i][j];
            s1[i] += e[i][j];
            s2[j] += e[i][j];
        }
    }
    
    int x = 1, y = 1;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == 'R') {
            e[x][y] = -s2[y];
            s1[x] -= s2[y];
            y++;
        } else {
            e[x][y] = -s1[x];
            s2[y] -= s1[x];
            x++;
        }
    }
    
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (i == n && j == m) e[i][j] = -s2[m];
            cout << e[i][j] << " ";
        }
        cout << endl;
    }
    
}

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