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HDU 1207 汉诺塔II (简单DP)
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; }
HDU 1207 汉诺塔II (简单DP)
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