HDU 1207 汉诺塔II (简单DP)

Posted 2020-08-27 dwtfukgv

题意：中文题。

析：在没有第四个柱子时，把 n 个盘子搬到第 3 个柱子时，那么2 ^ n -1次，由于多了一根，不知道搬到第四个柱子多少根时是最优的，

所以 dp[i] 表示搬到第4个柱子 i 个盘子时，步数最少，dp[i] = min{ dp[j] + (1<<i-j) - 1}。

也可以找规律，多写几个就发现规律。

代码如下：

找规律：

#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>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#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 double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -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 bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

LL dp[70];

void init(){
    dp[1] = 1;   dp[2] = 3;  dp[3] = 5;
    int cnt = 3;
    for(int i = 4, j = 0; i < 65; ++i, ++j){
        if(j == cnt){ j = 0, ++cnt;  }
        dp[i] = dp[i-1] + (1LL<<cnt-1);
    }
}

int main(){
    init();
    while(cin >> n)  cout << dp[n] << endl;
    return 0;
}

 DP：

#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>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -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 bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

ULL dp[70];

void init(){
    memset(dp, INF, sizeof dp);
    dp[1] = 1;   dp[2] = 3;  dp[3] = 5;
    for(int i = 4; i < 65; ++i)
        for(int j = 1; j < i; ++j)
            dp[i] = min(dp[i], dp[j]*2LL+(1LL<<i-j)-1);
}

int main(){
    init();
    while(cin >> n)  cout << dp[n] << endl;
    return 0;
}