abc 283E 经典dp

最新推荐文章于 2025-08-04 19:56:00 发布

aroyal菜菜

最新推荐文章于 2025-08-04 19:56:00 发布

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文章标签： c++ 算法图论

本文链接：https://blog.youkuaiyun.com/m0_62848317/article/details/130436516

该代码实现了一个求解二维矩阵中最小反转次数的动态规划算法。通过状态转移，计算每一行是否反转及其相邻行的状态，最终得到最小操作数。若无法达到目标状态，则输出-1。

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题意：https://www.luogu.com.cn/problem/AT_abc283_e

思路：非常经典的dp，设 $dp[i][j][k]$ 为前i行第i行是否反转和第i+1行是否反转。

/*keep on going and never give up*/
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
#define int long long
typedef pair<int, int> pii;
#define lowbit(x) x&(-x)
#define endl '\n'
#define wk is zqx ta die
int a[2][1005][1005];
int dp[1005][2][2];
//int dx[4] = {0, 0, 1, -1};
//int dy[4] = {1, -1, 0, 0};
signed main() {
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> a[0][i][j];
			a[1][i][j] = 1 - a[0][i][j];
		}
	}
	for (int i = 0; i <= n + 1; i++) {
		a[0][i][0] = a[0][i][m + 1] = a[1][i][0] = a[1][i][m + 1] = 2;
	}
	for (int i = 0; i <= m + 1; i++) {
		a[0][0][i] = a[0][n + 1][i] = a[1][0][i] = a[1][n + 1][i] = 2;
	}
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= 1; j++) {
			for (int k = 0; k <= 1; k++) {
				dp[i][j][k] = 1e18;
			}
		}
	}
	dp[0][0][0] = 0;
	dp[0][0][1] = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= 1; j++) {
			for (int k = 0; k <= 1; k++) {
				for (int g = 0; g <= 1; g++) {
					int ok = 0;
					for (int f = 1; f <= m; f++) {
						int x = a[k][i][f];
						int y = a[j][i - 1][f];
						int z = a[g][i + 1][f];
						int z1 = a[k][i][f + 1];
						int z2 = a[k][i][f - 1];
						if (z1 != x && z2 != x && y != x && z != x) {
							ok = 1;
							break;
						}
					}
					if (ok == 1) {
						continue;
					} else {
						dp[i][k][g] = min(dp[i][k][g], dp[i - 1][j][k] + k);
					}
				}
			}
		}
	}
	int ans = min(dp[n][0][0], dp[n][1][0]);
	if (ans == 1e18) {
		cout << "-1" << endl;
	} else {
		cout << ans << endl;
	}
	return 0;
}