hihocoder 1290 DP

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题目利用DP思想，dp[i][j][k]表示robot跑到i行j列目前移动方向为k时，所需要的最小的flip。其中0 <= i <= N，0 <= j <= M，k = right/down
dp[i][j][right] 可由dp[i][j-1][right]右移和dp[i-1][j][down]下移两种情况得到：
 1. dp[i][j-1][right]右移一步，到dp[i][j][right]，若maze[i][j] == 'b'，则dp[i][j][right]=dp[i][j-1][right] +1； 
2. dp[i-1][j][down]下移时，需要改变方向，先下移一步到第i行j列，此时方向是向下的，要变为向右，需考虑maze[i+1][j]，若i+1==n(边界)或者maze[i+1][j] == 'b'，则表示向下行不通，便直接改变方向向右；若第i+1行j列可以走的话，需将其变为'b'，使其行不通，才能改变方向为向右。故可表示为以下公式：
dp[i][j][right] = min(dp[i][j-1][right], dp[i-1][j][down] + (i+1 < n && maze[i+1][j] != 'b')) + (maze[i][j] == 'b');
dp[i][j][down]同理如下；
dp[i][j][down] = min(dp[i-1][j][down], dp[i][j-1][right] + (j+1 < m && maze[i][j+1] != 'b')) + (maze[i][j] == 'b');

#include <iostream>#include <vector>#include <algorithm>#include<map>#include<vector>#include<queue>#include<string>#include<set>#include<cmath>#include<cstdio>#include<sstream>#include<cstring>//#pragma warning(disable:4996)using namespace std;#define MAXN 50001#define INF INT_MAX#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))#define L(a) ((a)<<1)#define R(a) (((a)<<1)+1)int n, m;int dp[105][105][2];char a[105][105];int cal() {//0表示right ,1 表示downdp[0][0][0] = (a[0][0] == 'b');dp[0][0][1] = dp[0][0][0] + (1 < m&&a[0][1] != 'b');for (int i = 1; i < n; ++i) {dp[i][0][1] = min(dp[i - 1][0][0] + (1 < m&&a[i - 1][1] != 'b'), dp[i - 1][0][1]) + (a[i][0] == 'b');dp[i][0][0] = dp[i][0][1] + (i + 1 < n&&a[i + 1][0] != 'b');}for (int i = 1; i < m; ++i) {dp[0][i][0] = min(dp[0][i - 1][0], dp[0][i - 1][1] + (1 < m&&a[1][i - 1] != 'b')) + (a[0][i] == 'b');dp[0][i][1] = dp[0][i][0] + (i + 1 < m&&a[0][i + 1] != 'b');}for (int i = 1; i < n; ++i) {for (int j = 1; j < m; ++j) {dp[i][j][0] = min(dp[i][j - 1][0], dp[i - 1][j][1] + (i + 1 < n && a[i + 1][j] != 'b')) + (a[i][j] == 'b');dp[i][j][1] = min(dp[i - 1][j][1], dp[i][j - 1][0] + (j + 1 < m && a[i][j + 1] != 'b')) + (a[i][j] == 'b');}}return min(dp[n - 1][m - 1][0], dp[n - 1][m - 1][1]);}int main(){while (cin >> n >> m) {for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {cin >> a[i][j];}}cout << cal() << endl;}return 0;}

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