汉诺塔（记录每种路径次数）

https://ac.nowcoder.com/acm/contest/3004/I

题意：输出汉诺塔移动过程中每一种移动的次数和移动总数。

如下
A->B:XX
A->C:XX
B->A:XX
B->C:XX
C->A:XX
C->B:XX
SUM:XX

解法：记忆化搜索，当前状态的可以由上一状态得到。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
const int maxn = 110;
int n , m , k , t;
struct node{
    ll ab , ac , ba , bc , ca , cb;
}dp[maxn];

bool vis[maxn];

node hanio(int n , int a,  int b , int c){
    if(vis[n]) return dp[n];
    node x = hanio(n-1 , a , c , b);
    dp[n].ab += x.ac;
    dp[n].ac += x.ab;
    dp[n].ba += x.ca;
    dp[n].ca += x.ba;
    dp[n].bc += x.cb;
    dp[n].cb += x.bc;

    dp[n].ac++;
    x = hanio(n-1 , b , a , c);
    dp[n].ab += x.ba;
    dp[n].ac += x.bc;
    dp[n].ba += x.ab;
    dp[n].ca += x.cb;
    dp[n].bc += x.ac;
    dp[n].cb += x.ca;

    vis[n] = 1;
    return dp[n];
}

int main()
{
    vis[1] = 1 ;
    dp[1].ab = dp[1].ba = dp[1].bc = dp[1].ca = dp[1].cb = 0;
    dp[1].ac = 1 ;
    int n ; cin >> n;
    hanio(n , 1 , 2 , 3);
    ll sum = (1ll<<n)-1ll;
    cout << "A->B:" << dp[n].ab << endl;
    cout << "A->C:" << dp[n].ac << endl;
    cout << "B->A:" << dp[n].ba << endl;
    cout << "B->C:" << dp[n].bc << endl;
    cout << "C->A:" << dp[n].ca << endl;
    cout << "C->B:" << dp[n].cb << endl;
    cout << "SUM:" << sum << endl;
    return 0 ;
}

也可打表找规律。

原文地址：https://www.cnblogs.com/nonames/p/12293656.html