题目大意：有一个胖子在玩跳舞机，刚开始的位置在(0,0)，跳舞机有四个方向键，上左下右分别对应1,2,3,4.现在有以下规则
1.如果从0位置移动到任意四个位置，消耗能量2
2.如果从非0位置跳到相邻的位置，如1跳到2或4，消耗能量3
3.如果从非0位置跳到对面的位置，如2跳到4，消耗能量4
4.如果跳同一个位置，消耗能量1
5.两只脚不能在同一个位置

解题思路：这题其实很水，直接暴力就可以解决了，讨论所有情况，用dp[i][j][k]表示跳第k个数字，左脚在i这个位置，右脚在j这个位置时所消耗的能量，接着分类讨论

1.如果其中一只脚在0上的情况
2.其中一只脚踩的数字和当前要跳的数字一样
3.两只脚踩的数字和当前的数字不一样

三种情况，分别在细分即可，具体看代码

#include
   
     #include
    
      #include
     
       using namespace std; #define maxn 50010 #define INF 0x3f3f3f3f int dp[5][5][maxn]; int seq[maxn]; int strength[2] = {4,3}; int n; int solve() { memset(dp, 0x3f, sizeof(dp)); dp[0][seq[0]][0] = dp[seq[0]][0][0] = 2; for(int i = 1; i < n; i++) { for(int j = 0; j < 5; j++) { if(dp[j][seq[i-1]][i-1] != INF) { if(j == 0) { if(seq[i] != seq[i-1]) dp[seq[i]][seq[i-1]][i] = dp[j][seq[i-1]][i-1] + 2; if(seq[i] == seq[i-1]) dp[j][seq[i-1]][i] = dp[j][seq[i-1]][i-1] + 1; else dp[j][seq[i]][i] = dp[j][seq[i-1]][i-1] + strength[(seq[i-1] + seq[i]) % 2]; } else if(j == seq[i] || seq[i-1] == seq[i]) dp[j][seq[i-1]][i] = min(dp[j][seq[i-1]][i],dp[j][seq[i-1]][i-1] + 1); else { dp[seq[i]][seq[i-1]][i] = min(dp[j][seq[i-1]][i-1] + strength[(j + seq[i]) % 2], dp[seq[i]][seq[i-1]][i]); dp[j][seq[i]][i] = min(dp[j][seq[i-1]][i-1] + strength[(seq[i-1] + seq[i] ) % 2], dp[j][seq[i]][i]); } } if(dp[seq[i-1]][j][i-1] != INF) { if(j == 0) { if(seq[i] != seq[i-1]) dp[seq[i]][seq[i-1]][i] = dp[seq[i-1]][j][i-1] + 2; if(seq[i] == seq[i-1]) dp[seq[i-1]][j][i] = dp[seq[i-1]][j][i-1] + 1; else dp[seq[i]][j][i] = dp[seq[i-1]][j][i-1] + strength[(seq[i-1] + seq[i]) % 2]; } if(j == seq[i] || seq[i-1] == seq[i]) dp[seq[i-1]][j][i] = min(dp[seq[i-1]][j][i],dp[seq[i-1]][j][i-1] + 1); else { dp[seq[i]][seq[i-1]][i] = min(dp[seq[i-1]][j][i-1] + strength[(j + seq[i]) % 2], dp[seq[i]][seq[i-1]][i]); dp[seq[i]][j][i] = min(dp[seq[i-1]][j][i-1] + strength[(seq[i-1] + seq[i] ) % 2], dp[seq[i]][j][i]); } } } } int ans = INF; for(int i = 0; i < 5; i++) ans = min(min(ans, dp[seq[n-1]][i][n-1]), dp[i][seq[n-1]][n-1]); return ans; } int main() { n = 0; while(scanf("%d", &seq[n]) != EOF && seq[n++]) { while(scanf("%d", &seq[n]) && seq[n]) n++; printf("%d\n", solve()); n = 0; } return 0; } 
     
    
   

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