递推DP UVA 1291 Dance Dance Revolution

 1 /*
 2     题意：给一串跳舞的动作，至少一只脚落到指定的位置，不同的走法有不同的体力消耗，问最小体力消费多少
 3     DP：dp[i][j][k] 表示前i个动作，当前状态(j, k)的最小消费，状态转移方程：(a[i], k) <- min (a[i-1], k) + cost
 4         以及(a[i-1], a[i]) <- min (a[i-1], k) + cost, (k, a[i])和(a[i], a[i-1])情况类似，最后再去最小值就行了
 5 */
 6 /************************************************
 7  * Author        :Running_Time
 8  * Created Time  :2015-8-15 14:31:31
 9  * File Name     :UVA_1291.cpp
10  ************************************************/
11 
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30 
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 1e4 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 int a[MAXN];
38 int dp[MAXN][5][5];
39 
40 int cal(int x, int y)   {
41     int ret;
42     if (x == y) ret = 1;
43     else    {
44         if (y == 0)    ret = 2;
45         else    {
46             if (abs (x - y) == 2)   {
47                 ret = 4;
48             }
49             else    ret = 3;
50         }
51     }
52     return ret;
53 }
54 
55 int main(void)    {     //UVA 1291 Dance Dance Revolution
56     int n = 0;
57     while (scanf ("%d", &a[++n]) == 1)  {
58         if (a[1] == 0)  break;
59         while (a[n] != 0)   {
60             scanf ("%d", &a[++n]);
61         }
62         n--;
63         memset (dp, INF, sizeof (dp));
64         dp[1][a[1]][0] = (a[1] == 0 ? 1 : 2);
65         dp[1][0][a[1]] = (a[1] == 0 ? 1 : 2);
66         for (int i=2; i<=n; ++i)    {
67             for (int j=0; j<=4; ++j)    {
68                 int c1 = cal (a[i], j);
69                 int c2 = cal (a[i], a[i-1]);
70                 int x = a[i], y = a[i-1];
71                 dp[i][x][y] = min (dp[i][x][y], dp[i-1][j][y] + c1);
72                 dp[i][j][x] = min (dp[i][j][x], dp[i-1][j][y] + c2);
73 
74                 dp[i][y][x] = min (dp[i][y][x], dp[i-1][y][j] + c1);
75                 dp[i][x][j] = min (dp[i][x][j], dp[i-1][y][j] + c2);
76             }
77         }
78 
79         int ans = INF;
80         for (int i=0; i<=4; ++i)    {
81             ans = min (ans, min (dp[n][i][a[n]], dp[n][a[n]][i]));
82         }
83         printf ("%d\n", ans);   n = 0;
84     }
85 
86     return 0;
87 }