UVA - 1631 Locker(密码锁)(dp---记忆化搜索)

题意：有一个n(n<=1000)位密码锁，每位都是0~9，可以循环旋转。每次可以让1~3个相邻数字同时往上或者往下转一格。输入初始状态和终止状态(长度不超过1000),问最少要转几次。

分析：

1、从左往右依次使各个数字与终止状态相同。

2、dp[cur][x1][x2][x3]表示当前研究数字为第cur位，x1为a[cur]，x2为a[cur + 1]，x3为a[cur + 2]，在当前状态下，使所有数字变成终止状态的最小旋转次数。

3、研究第cur位时，第cur+1位和第cur+2位也可以随之一起转，枚举当前所有的旋转情况，并分别讨论向上和向下旋转两种情况。

#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

const double eps = 1e-8;

inline int dcmp(double a, double b){

    if(fabs(a - b) < eps) return 0;

    return a > b ? 1 : -1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const int MOD = 1e9 + 7;

const double pi = acos(-1.0);

const int MAXN = 1000 + 10;

const int MAXT = 10000 + 10;

using namespace std;

char aa[MAXN], bb[MAXN];

int a[MAXN], b[MAXN];

int dp[MAXN][15][15][15];

int len;

int getUpStep(int st, int et){//向上翻转的步数，向上转数字变大

    return (et - st + 10) % 10;

}

int getDownStep(int st, int et){

    return (st - et + 10) % 10;

}

int upNowpos(int st, int length){

    return (st + length) % 10;

}

int downNowpos(int st, int length){//从st向下翻转length步变成的数字

    return (st - length + 10) % 10;

}

int dfs(int cur, int x1, int x2, int x3){

    if(dp[cur][x1][x2][x3] != -1) return dp[cur][x1][x2][x3];

    if(cur == len) return 0;

    int ans = INT_INF;

    int upstep = getUpStep(x1, b[cur]), downstep = getDownStep(x1, b[cur]);

    for(int i = 0; i <= upstep; ++i){//枚举第cur+1位可以跟着第cur位一起向上旋转的步数

        for(int j = 0; j <= i; ++j){//枚举第cur+2位可以跟着第cur位和第cur+1位一起向上旋转的步数

            ans = Min(ans, dfs(cur + 1, upNowpos(x2, i), upNowpos(x3, j), a[cur + 3]) + upstep);

        }

    }

    for(int i = 0; i <= downstep; ++i){//向下转

        for(int j = 0; j <= i; ++j){

            ans = Min(ans, dfs(cur + 1, downNowpos(x2, i), downNowpos(x3, j), a[cur + 3]) + downstep);

        }

    }

    return dp[cur][x1][x2][x3] = ans;

}

int main(){

    while(scanf("%s%s", aa, bb) == 2){

        memset(dp, -1, sizeof dp);

        len = strlen(aa);

        for(int i = 0; i < len; ++i) a[i] = aa[i] - '0';

        for(int i = 0; i < len; ++i) b[i] = bb[i] - '0';

        a[len] = a[len + 1] = a[len + 2] = 0;

        b[len] = b[len + 1] = b[len + 2] = 0;

        printf("%d\n", dfs(0, a[0], a[1], a[2]));

    }

    return 0;

}