hdu 1195 Open the Lock

Problem - 1195

　　一个全排列的搜索题，开始的时候看错题，没发觉是两个相邻的数才可以交换。计算交换次数可以用冒泡排序来模拟，从而算两个状态间要多少次交换才能达到。

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); i++)
#define REP_1(i, n) for (int i = 1; i <= (n); i++)
#define INC(i, a, b) for (int i = (a); i <= (b); i++)
#define DEC(i, a, b) for (int i = (a); i >= (b); i--)
#define _clr(x) memset(x, 0, sizeof(x))

template <class T> T sqr(T x) {
    return x * x;
}
typedef long long LL;
const int N = 1e5 + 100;
const LL linf = 0x5555555555555555ll;
const int inf = 0x55555555;

int arr[5], best;
char str[2][6];

int diff(char a, char b) {
    int d = max(a, b) - min(a, b);
    return min(d, 9 - d);
}

int cal() {
    int ret = 0;
    REP(i, 4) ret += diff(str[0][arr[i]], str[1][i]);
    return ret;
}

int cal(int *a) {
    int tmp[4];
    REP(i, 4) tmp[i] = a[i];
    int ret = 0;
    bool chg = true;
    while (chg) {
        chg = false;
        REP(i, 3) {
            if (tmp[i] > tmp[i + 1]) { swap(tmp[i], tmp[i + 1]); chg = true; ret++;}
        }
    }
    return ret;
}

void dfs(int l, int r) {
    if (l == r) {
        best = min(best, cal() + cal(arr));
        return ;
    }
    INC(i, l, r) {
        swap(arr[i], arr[l]);
        dfs(l + 1, r);
        swap(arr[i], arr[l]);
    }
}

int main() {
//    freopen("in", "r", stdin);
    int T;
    cin >> T;
    while (T--) {
        REP(i, 2) cin >> str[i];
        REP(i, 4) arr[i] = i;
        best = inf;
        dfs(0, 3);
        cout << best << endl;
    }
    return 0;
}

——written by Lyon