UVa 1631 Locker (DP)

题意：有一个 n 位密码锁，每位都是0-9，可以循环旋转。同时可以让1-3个相邻数字进行旋转一个，给定初始状态和目状态，问你最少要转多少次。

析：很明显的一个DP题。dp[i][j][k] 表示前 i 位已经转好，并且第 i+1 位是 j ，第 i+2 位是 k，那么我们先把第 i 位转到指定位置，然后计算转多少次，

然后再考虑 i+1位和 i+2位，要旋转小于等于第 i 位的次数，这就转移完了。比较简单的一个DP，只是没有遇见过。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s1[maxn], s2[maxn];
int a[maxn], b[maxn];
int g[2][15][15];
LL dp[maxn][15][15];

int solve(int pre, int last, int cnt){
    return cnt ? (last+10-pre) % 10 : (pre+10-last) % 10;
}

int main(){
    for(int i = 0; i < 2; ++i)  for(int j = 0; j < 10; ++j)
        for(int k = 0; k < 10; ++k)  g[i][j][k] = i ? (j+k) % 10 : (j-k+10) % 10;

    while(scanf("%s %s", s1, s2) == 2){
        n = strlen(s1);
        for(int i = 1; i <= n; ++i) a[i] = s1[i-1] - ‘0‘, b[i] = s2[i-1] - ‘0‘;
        a[n+1] = a[n+2] = b[n+1] = b[n+2] = 0;
        for(int i = 0; i <= n; ++i) for(int j = 0; j < 10; ++j)
            for(int k = 0; k < 10; ++k) dp[i][j][k] = LNF;

        dp[0][a[1]][a[2]] = 0;
        for(int i = 1; i <= n; ++i){
            for(int r = 0; r < 2; ++r){
                for(int j = 0; j < 10; ++j){
                    for(int k = 0; k < 10; ++k){
                        int x = solve(j, b[i], r);
                        for(int ii = 0; ii <= x; ++ii){
                            for(int jj = 0; jj <= ii; ++jj){
                                dp[i][g[r][k][ii]][g[r][a[i+2]][jj]] = Min(dp[i][g[r][k][ii]][g[r][a[i+2]][jj]], dp[i-1][j][k] + x);
                            }
                        }
                    }
                }
            }
        }

        printf("%lld\n", dp[n][0][0]);
    }
    return 0;
}