POJ 3278 Catch That Cow
Posted zaq19970105
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 3278 Catch That Cow相关的知识,希望对你有一定的参考价值。
题目链接:https://vjudge.net/problem/POJ-3278
题目大意
给定 N,K ,N 每次可以 +1,-1,*2,问 N 变到 K 最少要几次变化?
分析
一开始用 DP 死活做不出来,然后我去打了一周目魂2,再回来看看这道题。然后发现这是道 bfs,特水的那种。。。
代码如下
1 #include <cmath> 2 #include <ctime> 3 #include <iostream> 4 #include <string> 5 #include <vector> 6 #include <cstdio> 7 #include <cstdlib> 8 #include <cstring> 9 #include <queue> 10 #include <map> 11 #include <set> 12 #include <algorithm> 13 #include <cctype> 14 #include <stack> 15 #include <deque> 16 #include <list> 17 #include <sstream> 18 using namespace std; 19 20 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 21 #define Rep(i,n) for (int i = 0; i < (n); ++i) 22 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 23 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 24 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 25 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 26 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 27 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 28 29 #define pr(x) cout << #x << " = " << x << " " 30 #define prln(x) cout << #x << " = " << x << endl 31 32 #define LOWBIT(x) ((x)&(-x)) 33 34 #define ALL(x) x.begin(),x.end() 35 #define INS(x) inserter(x,x.begin()) 36 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 37 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 38 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 39 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 40 41 #define ms0(a) memset(a,0,sizeof(a)) 42 #define msI(a) memset(a,inf,sizeof(a)) 43 #define msM(a) memset(a,-1,sizeof(a)) 44 45 #define MP make_pair 46 #define PB push_back 47 #define ft first 48 #define sd second 49 50 template<typename T1, typename T2> 51 istream &operator>>(istream &in, pair<T1, T2> &p) 52 in >> p.first >> p.second; 53 return in; 54 55 56 template<typename T> 57 istream &operator>>(istream &in, vector<T> &v) 58 for (auto &x: v) 59 in >> x; 60 return in; 61 62 63 template<typename T1, typename T2> 64 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) 65 out << "[" << p.first << ", " << p.second << "]" << "\n"; 66 return out; 67 68 69 inline int gc() 70 static const int BUF = 1e7; 71 static char buf[BUF], *bg = buf + BUF, *ed = bg; 72 73 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 74 return *bg++; 75 76 77 inline int ri() 78 int x = 0, f = 1, c = gc(); 79 for(; c<48||c>57; f = c==‘-‘?-1:f, c=gc()); 80 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 81 return x*f; 82 83 84 template<class T> 85 inline string toString(T x) 86 ostringstream sout; 87 sout << x; 88 return sout.str(); 89 90 91 inline int toInt(string s) 92 int v; 93 istringstream sin(s); 94 sin >> v; 95 return v; 96 97 98 //min <= aim <= max 99 template<typename T> 100 inline bool BETWEEN(const T aim, const T min, const T max) 101 return min <= aim && aim <= max; 102 103 104 typedef long long LL; 105 typedef unsigned long long uLL; 106 typedef pair< double, double > PDD; 107 typedef pair< int, int > PII; 108 typedef pair< int, PII > PIPII; 109 typedef pair< string, int > PSI; 110 typedef pair< int, PSI > PIPSI; 111 typedef set< int > SI; 112 typedef set< PII > SPII; 113 typedef vector< int > VI; 114 typedef vector< double > VD; 115 typedef vector< VI > VVI; 116 typedef vector< SI > VSI; 117 typedef vector< PII > VPII; 118 typedef map< int, int > MII; 119 typedef map< int, string > MIS; 120 typedef map< int, PII > MIPII; 121 typedef map< PII, int > MPIII; 122 typedef map< string, int > MSI; 123 typedef map< string, string > MSS; 124 typedef map< PII, string > MPIIS; 125 typedef map< PII, PII > MPIIPII; 126 typedef multimap< int, int > MMII; 127 typedef multimap< string, int > MMSI; 128 //typedef unordered_map< int, int > uMII; 129 typedef pair< LL, LL > PLL; 130 typedef vector< LL > VL; 131 typedef vector< VL > VVL; 132 typedef priority_queue< int > PQIMax; 133 typedef priority_queue< int, VI, greater< int > > PQIMin; 134 const double EPS = 1e-8; 135 const LL inf = 0x3fffffff; 136 const LL infLL = 0x3fffffffffffffffLL; 137 const LL mod = 1e9 + 7; 138 const int maxN = 1e5 + 7; 139 const LL ONE = 1; 140 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 141 const LL oddBits = 0x5555555555555555; 142 143 int N, K, ans; 144 queue< int > Q; 145 bool vis[maxN]; 146 147 void bfs() 148 int cnt = 1; 149 Q.push(N); 150 151 while(!Q.empty()) 152 int tmp = Q.front(); Q.pop(); 153 --cnt; 154 if(tmp == K) break; 155 156 if(BETWEEN(tmp, 0, maxN) && !vis[tmp]) 157 vis[tmp] = true; 158 Q.push(tmp - 1); 159 Q.push(tmp + 1); 160 Q.push(tmp * 2); 161 162 163 if(cnt == 0) 164 cnt = Q.size(); 165 ++ans; 166 167 168 169 170 int main() 171 //freopen("MyOutput.txt","w",stdout); 172 //freopen("input.txt","r",stdin); 173 INIT(); 174 cin >> N >> K; 175 bfs(); 176 cout << ans << endl; 177 return 0; 178
以上是关于POJ 3278 Catch That Cow的主要内容,如果未能解决你的问题,请参考以下文章
POJ 3278 - Catch That Cow - [BFS]