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递推DP UVA 1291 Dance Dance Revolution

2015-08-15 15:33 375 查看
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
!= 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
[i][a
], dp
[a
][i]));
}
printf ("%d\n", ans);   n = 0;
}

return 0;
}


  
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