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

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< string, int > PSI;
typedef set< int > SI;
typedef vector< int > VI;
typedef vector< VI > VVI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef multimap< int, int > MMII;
typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int n;
VVI block; 
MII pos;

// 把 a 号块上面的所有方块复位 
void reSet(int a) {
    int x;
    while((x = block[pos[a]].back()) != a) {
        block[pos[a]].pop_back();
        block[x].PB(x);
        pos[x] = x;
    }
}

// pile a over b
void D(int a, int b) {
    stack< int > sk;
    int x;
    do {
        x = block[pos[a]].back();
        block[pos[a]].pop_back();
        sk.push(x);
        pos[x] = pos[b];
    }while(x != a);
    
    while(!sk.empty()) {
        block[pos[b]].PB(sk.top());
        sk.pop();
    }
}

// move a onto b
void A(int a, int b) {
    reSet(a);
    reSet(b);
    D(a, b);
}

// move a over b
void B(int a, int b) {
    reSet(a);
    D(a, b);
}

// pile a onto b
void C(int a, int b) {
    reSet(b);
    D(a, b);
}

void (*func[4])(int, int) = {A, B, C, D};

int main(){
    INIT();
    cin >> n;
    block.assign(n, VI());
    Rep(i, n) {
        block[i].PB(i);
        pos[i] = i;
    }
    
    string tmp;
    int x, y, mode;
    while(cin >> tmp) {
        mode = 0;
        if(tmp[0] == 'q') break;
        if(tmp[0] == 'p') mode |= 2;
        cin >> x >> tmp >> y;
        if(tmp[1] == 'v') mode |= 1;
        if(pos[x] == pos[y]) continue;
        func[mode](x, y);
    }
    
    Rep(i, n) {
        cout << i << ":";
        foreach(j, block[i]) cout << " " << *j;
        cout << endl;
    }
    return 0;
}