• hdu 1430 魔板 康托展开 + 很好的映射


    http://acm.hdu.edu.cn/showproblem.php?pid=1430

    如果从start ---> end,每一次都bfs进行,那么就肯定会超时。

    考虑到先把start映射到原始状态"12345678",然后又把end按照同样的规则,就是start的变化,映射到某一个地方。那么就可以看作是从固定的起点“12345678”-->newend,这个就可以打表了。

    比如从87654321变化到12345678,那么就是8变成了1,7变成了2,6变成了3....。。那么end那里也按照这样变,在end中找一个8,变成1......以此类推。

    相当于模仿start走动而已,肯定能走到的。

    然后就可以打表。

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <assert.h>
    #define IOS ios::sync_with_stdio(false)
    using namespace std;
    #define inf (0x3f3f3f3f)
    typedef long long int LL;
    
    
    #include <iostream>
    #include <sstream>
    #include <vector>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #include <bitset>
    
    class cantor {  //求1...n的排列的第k大 && hash排列,
    public : ///class默认是private,所以要加上public
        int fac[12];
        cantor() { //预处理阶乘
            fac[0] = 1;
            for (int i = 1; i <= 11; ++i) {
                fac[i] = fac[i - 1] * i;
            }
        }
        int decode(char str[], int lenstr) { //O(n * n)的hash
            int ans = 0;
            for (int i = 1; i < lenstr; ++i) {
                int cnt = 0;
                for (int j = i + 1; j <= lenstr; ++j) {
                    if (str[i] > str[j]) {
                        cnt++;
                    }
                }
                ans += cnt * fac[lenstr - i];
            }
            return ans + 1;
        }
        vector<int> encode(int lenstr, int k) { //字典序排第k的是那个,k从1开始
            vector<int>ans;
            int toans;
            bool vis[12] = {0};
            for (int i = 1; i <= lenstr; ++i) {
                int t = k / fac[lenstr - i];
                k %= fac[lenstr - i];
                for (toans = 1; toans <= lenstr; ++toans) {
                    if (vis[toans]) continue;
                    if (t == 0) break;
                    t--;
                }
                ans.push_back(toans);
                vis[toans] = true;
            }
            return ans;
        }
    } cantor;
    char be[23], en[23];
    void A(char str[]) {
        for (int i = 1; i <= 4; ++i) {
            swap(str[i], str[9 - i]);
        }
    }
    void B(char str[]) {
        char t = str[4];
        for (int i = 4; i >= 2; --i) str[i] = str[i - 1];
        str[1] = t;
        t = str[5];
        for (int i = 5; i < 8; ++i) str[i] = str[i + 1];
        str[8] = t;
    }
    void C(char str[]) {
        char t1 = str[2];
        str[2] = str[7];
        char t2 = str[3];
        str[3] = t1;
        t1 = str[6];
        str[6] = t2;
        str[7] = t1;
    }
    struct node {
        char str[12];
        string now;
        node(char ss[]) {
            strcpy(str + 1, ss + 1);
        }
        node() {}
    } que[40320 + 20];
    bool vis[40320 + 20];
    string ans[40320 + 20];
    void bfs() {
        char st[55] = "q12345678";
        int fr, tail;
        fr = tail = 0;
        que[tail++] = node(st);
        que[fr].now = "";
        vis[cantor.decode(st, 8)] = true;
        while (fr < tail) {
            char t[22];
            strcpy(t + 1, que[fr].str + 1);
            A(t);
            int valt = cantor.decode(t, 8);
            if (vis[valt] == false) {
                vis[valt] = true;
                strcpy(que[tail].str + 1, t + 1);
                que[tail].now = que[fr].now + "A";
                ans[valt] = que[tail].now;
                tail++;
            }
    
            strcpy(t + 1, que[fr].str + 1);
            B(t);
            valt = cantor.decode(t, 8);
            if (vis[valt] == false) {
                vis[valt] = true;
                strcpy(que[tail].str + 1, t + 1);
                que[tail].now = que[fr].now + "B";
                ans[valt] = que[tail].now;
                tail++;
            }
    
            strcpy(t + 1, que[fr].str + 1);
            C(t);
            valt = cantor.decode(t, 8);
            if (vis[valt] == false) {
                vis[valt] = true;
                strcpy(que[tail].str + 1, t + 1);
                que[tail].now = que[fr].now + "C";
                ans[valt] = que[tail].now;
                tail++;
            }
            fr++;
        }
    }
    int pos[22], f[22];
    char ten[22];
    void work() {
        for (int i = 1; i <= 8; ++i) {
            pos[be[i] - '0'] = i;
        }
        for (int i = 1; i <= 8; ++i) {
            en[i] = pos[en[i] - '0'] + '0';
        }
    //    cout << en + 1 << endl;
        cout << ans[cantor.decode(en, 8)] << endl;
    }
    int main() {
    #ifdef local
        freopen("data.txt", "r", stdin);
    //    freopen("data.txt", "w", stdout);
    #endif
    //    cout << cantor.decode("q87645321", 8) << endl;
        bfs();
        while (scanf("%s%s", be + 1, en + 1) != EOF) work();
        return 0;
    }
    View Code

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  • 原文地址:https://www.cnblogs.com/liuweimingcprogram/p/6402305.html
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