4 切割纸片

作者: 赵晓鹏时间限制: 1S章节: 动态规划与贪心

---------------------------------输入

6 4
1 1 0 1
0 0 0 1
0 0 0 1
0 0 0 1
0 0 1 1
0 0 1 1

--------------------输出结果

4
 

#include <iostream>
#include <vector>
#include <climits>using namespace std;int minCut(vector<vector<int>>& paper, int startRow, int endRow, int startCol, int endCol, vector<vector<vector<vector<int>>>>& dp) {// 如果已经计算过该区域的最小切割数，直接返回结果if (dp[startRow][endRow][startCol][endCol] != -1) {return dp[startRow][endRow][startCol][endCol];}// 检查当前区域是否是全带洞的纸片或完整的纸片bool allHoles = true;bool allIntact = true;for (int i = startRow; i < endRow; i++) {for (int j = startCol; j < endCol; j++) {if (paper[i][j] == 1) {allIntact = false;}else {allHoles = false;}if (!allHoles && !allIntact) {break;}}if (!allHoles && !allIntact) {break;}}// 如果是全带洞的纸片或完整的纸片，则返回0if (allHoles || allIntact) {dp[startRow][endRow][startCol][endCol] = 0;return 0;}int minCuts = INT_MAX;// 水平切割for (int i = startRow + 1; i < endRow; i++) {int cuts = minCut(paper, startRow, i, startCol, endCol, dp) + minCut(paper, i, endRow, startCol, endCol, dp) + 1;minCuts = min(minCuts, cuts);}// 垂直切割for (int j = startCol + 1; j < endCol; j++) {int cuts = minCut(paper, startRow, endRow, startCol, j, dp) + minCut(paper, startRow, endRow, j, endCol, dp) + 1;minCuts = min(minCuts, cuts);}dp[startRow][endRow][startCol][endCol] = minCuts;return minCuts;
}int main() {int m, n;cin >> m >> n;vector<vector<int>> paper(m, vector<int>(n));for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {cin >> paper[i][j];}}// 初始化dp数组为-1vector<vector<vector<vector<int>>>> dp(m, vector<vector<vector<int>>>(m + 1, vector<vector<int>>(n, vector<int>(n + 1, -1))));int minCuts = minCut(paper, 0, m, 0, n, dp);cout << minCuts << endl;return 0;
}