题解 P1242 【新汉诺塔】

MisTH · 2018-02-03 21:04:44 · 题解

广大人名群众所喜闻乐见的汉诺塔。

我的思路是这样的：将某个圆片v从a移到b，必须要先将所有比v小的圆片从a、b移到c，所以从v-1开始从大到小依次把所有圆片移到c上，这就可以抽象出递归过程了:

void dfs(int v, int from, int to, int by) { // 将圆片v从from移到to
    if (from == to)return;
    for (int i = v - 1; i > 0; i--) { // 先将所有比v小的圆片移到by
        dfs(i, pos[i], by, 3 - pos[i] - by);
    }
    mov(v, from, to);
}

易证这样得到的是最简方案。

完整代码：

#include<iostream>

using namespace std;

int stack[3][66];
int p[3];
int tar[66];
int pos[66];
int ans = 0;

void mov(int v, int from, int to) {
    cout << "move " << v << " from " << (char) (from + 'A') << " to " << (char) (to + 'A') << endl;
    p[from]--;
    stack[to][p[to]++] = v;
    pos[v] = to;
    ans++;
}

void dfs(int v, int from, int to, int by) { // 将圆片v从from移到to
    if (from == to)return;
    for (int i = v - 1; i > 0; i--) { // 先将所有比v小的圆片移到by
        dfs(i, pos[i], by, 3 - pos[i] - by);
    }
    mov(v, from, to);
}

int main() {
    ios::sync_with_stdio(false);
    int n, m, t;
    cin >> n;
    for (int i = 0; i < 3; i++) {
        cin >> m;
        p[i] = m + 1;
        stack[i][0] = 66;
        for (int j = 1; j <= m; j++) {
            cin >> stack[i][j];
            pos[stack[i][j]] = i;
        }
    }
    for (int i = 0; i < 3; i++) {
        cin >> m;
        for (int j = 0; j < m; j++) {
            cin >> t;
            tar[t] = i;
        }
    }
    for (int i = n; i > 0; i--) {
        if (pos[i] != tar[i]) {
            dfs(i, pos[i], tar[i], 3 - pos[i] - tar[i]);
        }
    }
    cout << ans;
}

风格略丑，神犇们请轻喷orz