CF 1689 D. Lena and Matrix 思维 1900

CF 1689 D. Lena and Matrix 思维 1900

题意：一个 n x m 的矩阵，有白点和黑点，找到一个任意颜色的点，使他到最远的黑点距离最小。

思路：最远的黑点一定在四个角，先找到这四个点，再遍历所有点和这四个点计算。找到最小的最远点。左上角的点是 i + j 最小，右上角的点是 i + （-j） 最小，左下角的点是 -i + j 最小，右下角的点是 i + j 最大。

代码：

#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int N = 2e5 + 10, P = 1e9 + 7, mod = 998244353;
char a[1001][1001];
PII b[4];
void solve(){int n, m;cin >> n >> m;memset(b, 0, sizeof b);for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {cin >> a[i][j];if(a[i][j] == 'W') continue;if(!b[0].first || i + j < b[0].first + b[0].second)b[0] = {i, j};if(!b[1].first || i - j < b[1].first - b[1].second)b[1] = {i, j};if(!b[2].first || i - j > b[2].first - b[2].second)b[2] = {i, j};if(!b[3].first || i + j > b[3].first + b[3].second)b[3] = {i, j};}}int minn = 0x3f3f3f3f;PII ans;for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++){int res = 0;for(int k = 0; k < 4; k++)res = max(res, abs(i - b[k].first) + abs(j - b[k].second));if(res < minn) minn = res, ans = {i, j};}}cout << ans.first << " " << ans.second << endl;
}                 
int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);int t;cin >> t;while(t--) {solve();}return 0;
}