标签：namespace tar queue ret span ref include name int
一个n*m的迷宫由‘.’和‘b‘组成，从(1,1)走到(n,m)，只能向右或者向下走，但遇到‘b’时才能改变方向，开始时方向向右。
问到达(n,m)至少改变几个位置上的值
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原来转移方程也可以这么优美
每个方格有两个状态，向右和向下
这两个状态均由左边和上边的两个方格四个状态转移得来
#include <set> #include <map> #include <stack> #include <queue> #include <cmath> #include <vector> #include <string> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #define MAX(a,b) ((a)>=(b)?(a):(b)) #define MIN(a,b) ((a)<=(b)?(a):(b)) #define long long LL #define OO 0x0fffffff using namespace std; #define RIGHT 0 #define DOWN 1 const int N = 111; char maze[N][N]; int dp[N][N][2]; int main(){ int n,m; cin>>n>>m; for(int i=0;i<n;i++) cin>>maze[i]; for(int i=0;i<n;i++) maze[i][m] = ‘b‘; for(int j=0;j<m;j++) maze[n][j] = ‘b‘; dp[0][0][0] = (maze[0][0]==‘b‘); dp[0][0][1] = (maze[0][0]==‘b‘)+(maze[0][1]!=‘b‘); for(int i=0;i<n;i++) for(int j=0;j<m;j++){ if(!(i+j)) continue; dp[i][j][0] = dp[i][j][1] = OO; if(j-1>=0){ dp[i][j][0] = MIN(dp[i][j-1][0],dp[i][j-1][1]+(maze[i+1][j-1]!=‘b‘)); dp[i][j][1] = MIN(dp[i][j-1][0]+(maze[i][j+1]!=‘b‘),dp[i][j-1][1]+(maze[i+1][j-1]!=‘b‘)+(maze[i][j+1]!=‘b‘)); } if(i-1>=0){ dp[i][j][0] = MIN(dp[i][j][0],dp[i-1][j][0]+(maze[i-1][j+1]!=‘b‘)+(maze[i+1][j]!=‘b‘)); dp[i][j][0] = MIN(dp[i][j][0],dp[i-1][j][1]+(maze[i+1][j]!=‘b‘)); dp[i][j][1] = MIN(dp[i][j][1],dp[i-1][j][0]+(maze[i-1][j+1]!=‘b‘)); dp[i][j][1] = MIN(dp[i][j][1],dp[i-1][j][1]); } dp[i][j][0]+=(maze[i][j]!=‘.‘); dp[i][j][1]+=(maze[i][j]!=‘.‘); } printf("%d\n",MIN(dp[n-1][m-1][0],dp[n-1][m-1][1])); return 0; }
标签：namespace tar queue ret span ref include name int
原文地址：http://www.cnblogs.com/redips-l/p/6852579.html
