标签：else name col ORC test ace 注意 oid clu
题目链接：https://codeforces.com/contest/1359/problem/B
因为 $1 \times 2$ 的瓷砖不能旋转，所以每次逐行考虑即可，注意 $y$ 取 $min(2x, y)$ 。
#include <bits/stdc++.h> using namespace std; void solve() { int n, m, x, y; cin >> n >> m >> x >> y; y = min(2 * x, y); int ans = 0; for (int i = 0; i < n; i++) { string s; cin >> s; for (int j = 0; j < m; j++) { if (s[j] == ‘.‘) { if (j + 1 < m and s[j + 1] == ‘.‘) ans += y, ++j; else ans += x; } } } cout << ans << "\n"; } int main() { int t; cin >> t; while (t--) solve(); }
Educational Codeforces Round 88 (Rated for Div. 2) B. New Theatre Square
标签：else name col ORC test ace 注意 oid clu
原文地址：https://www.cnblogs.com/Kanoon/p/12989137.html
