递推DP UVA 1291 Dance Dance Revolution

 

题目传送门

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
分析：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

收获：四个状态转移方向

 

代码：

/************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-15 14:31:31
 * File Name     :UVA_1291.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[MAXN];
int dp[MAXN][5][5];

int cal(int x, int y)   {
    int ret;
    if (x == y) ret = 1;
    else    {
        if (y == 0)    ret = 2;
        else    {
            if (abs (x - y) == 2)   {
                ret = 4;
            }
            else    ret = 3;
        }
    }
    return ret;
}

int main(void)    {     //UVA 1291 Dance Dance Revolution
    int n = 0;
    while (scanf ("%d", &a[++n]) == 1)  {
        if (a[1] == 0)  break;
        while (a[n] != 0)   {
            scanf ("%d", &a[++n]);
        }
        n--;
        memset (dp, INF, sizeof (dp));
        dp[1][a[1]][0] = (a[1] == 0 ? 1 : 2);
        dp[1][0][a[1]] = (a[1] == 0 ? 1 : 2);
        for (int i=2; i<=n; ++i)    {
            for (int j=0; j<=4; ++j)    {
                int c1 = cal (a[i], j);
                int c2 = cal (a[i], a[i-1]);
                int x = a[i], y = a[i-1];
                dp[i][x][y] = min (dp[i][x][y], dp[i-1][j][y] + c1);
                dp[i][j][x] = min (dp[i][j][x], dp[i-1][j][y] + c2);

                dp[i][y][x] = min (dp[i][y][x], dp[i-1][y][j] + c1);
                dp[i][x][j] = min (dp[i][x][j], dp[i-1][y][j] + c2);
            }
        }

        int ans = INF;
        for (int i=0; i<=4; ++i)    {
            ans = min (ans, min (dp[n][i][a[n]], dp[n][a[n]][i]));
        }
        printf ("%d\n", ans);   n = 0;
    }

    return 0;
}