UVA 101 The Blocks Problem

题目链接：https://vjudge.net/problem/UVA-101

题目大意

　　初始时从左到右有n个木块，编号为0~n-1,要求实现下列四种操作:

1. move a onto b: 把a和b上方的木块全部放回初始的位置，然后把a放到b上面
2. move a over b: 把a上方的木块全部放回初始的位置，然后把a放在b所在木块堆的最上方
3. pile a onto b: 把b上方的木块部放回初始的位置，然后把a和a上面所有的木块整体放到b上面
4. pile a over b: 把a和a上面所有的木块整体放在b所在木块堆的最上方

　　一组数据的结束标志为"quit"，如果有非法指令(a和b在同一堆)，应当忽略。

　　输出:所有操作输入完毕后，从左到右，从下到上输出每个位置的木块编号。

　　翻译搬运自洛谷。

分析

　　一道简单的模拟题，没什么技术含量，纯Coding。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "\n";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< string, int > PSI;
 69 typedef set< int > SI;
 70 typedef vector< int > VI;
 71 typedef vector< VI > VVI;
 72 typedef vector< PII > VPII;
 73 typedef map< int, int > MII;
 74 typedef multimap< int, int > MMII;
 75 typedef unordered_map< int, int > uMII;
 76 typedef pair< LL, LL > PLL;
 77 typedef vector< LL > VL;
 78 typedef vector< VL > VVL;
 79 typedef priority_queue< int > PQIMax;
 80 typedef priority_queue< int, VI, greater< int > > PQIMin;
 81 const double EPS = 1e-10;
 82 const LL inf = 0x7fffffff;
 83 const LL infLL = 0x7fffffffffffffffLL;
 84 const LL mod = 1e9 + 7;
 85 const int maxN = 1e5 + 7;
 86 const LL ONE = 1;
 87 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 88 const LL oddBits = 0x5555555555555555;
 89 
 90 int n;
 91 VVI block; 
 92 MII pos;
 93 
 94 // 把 a 号块上面的所有方块复位 
 95 void reSet(int a) {
 96     int x;
 97     while((x = block[pos[a]].back()) != a) {
 98         block[pos[a]].pop_back();
 99         block[x].PB(x);
100         pos[x] = x;
101     }
102 }
103 
104 // pile a over b
105 void D(int a, int b) {
106     stack< int > sk;
107     int x;
108     do {
109         x = block[pos[a]].back();
110         block[pos[a]].pop_back();
111         sk.push(x);
112         pos[x] = pos[b];
113     }while(x != a);
114     
115     while(!sk.empty()) {
116         block[pos[b]].PB(sk.top());
117         sk.pop();
118     }
119 }
120 
121 // move a onto b
122 void A(int a, int b) {
123     reSet(a);
124     reSet(b);
125     D(a, b);
126 }
127 
128 // move a over b
129 void B(int a, int b) {
130     reSet(a);
131     D(a, b);
132 }
133 
134 // pile a onto b
135 void C(int a, int b) {
136     reSet(b);
137     D(a, b);
138 }
139 
140 void (*func[4])(int, int) = {A, B, C, D};
141 
142 int main(){
143     INIT();
144     cin >> n;
145     block.assign(n, VI());
146     Rep(i, n) {
147         block[i].PB(i);
148         pos[i] = i;
149     }
150     
151     string tmp;
152     int x, y, mode;
153     while(cin >> tmp) {
154         mode = 0;
155         if(tmp[0] == 'q') break;
156         if(tmp[0] == 'p') mode |= 2;
157         cin >> x >> tmp >> y;
158         if(tmp[1] == 'v') mode |= 1;
159         if(pos[x] == pos[y]) continue;
160         func[mode](x, y);
161     }
162     
163     Rep(i, n) {
164         cout << i << ":";
165         foreach(j, block[i]) cout << " " << *j;
166         cout << endl;
167     }
168     return 0;
169 }

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