The Blocks Problem （vector入门题）

Sample Input

10

move 9 onto 1

move 8 over 1

move 7 over 1

move 6 over 1

pile 8 over 6

pile 8 over 5

move 2 over 1

move 4 over 9

quit

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Sample Output

0: 0

1: 1 9 2 4

2:

3: 3

4:

5: 5 8 7 6

6:

7:

8:

9:

题目大意是通过指定的4种指令来改变木块的位置。用vector创个数组，分解4个指令并封装方法即可。

贴代码：

#include <iostream>
#include <string>
#include <sstream>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
using namespace std;
#define Start clock_t start = clock();
#define End clock_t end = clock();
#define Print cout<<(double)(end-start)/CLOCKS_PER_SEC*1000<<"毫秒"<<endl;

int n;
const int maxn = 30;
vector<int> pile[maxn];

void find_block(int a,int& p,int& h){
    for(p = 0;p < n;p++){
        for(h = 0;h < pile[p].size();h++){
            if(a == pile[p][h])return;
        }
    }
}

void clear_block(int a,int p,int h){
    for(int i = h+1;i < pile[p].size();i++){
        int k = pile[p][i];
        pile[k].push_back(k);
    }
    pile[p].resize(h+1);
}

void pile_on(int a,int p,int h,int p2){
    for(int i = h;i < pile[p].size();i++){
        pile[p2].push_back(pile[p][i]);
    }
    pile[p].resize(h);
}

void print(){
    for(int i = 0;i < n;i++){
        printf("%d:",i);
        for(int j = 0;j < pile[i].size();j++){
            printf(" %d",pile[i][j]);
        }
        printf("\n");
    }
}
int main()
{
    //Start;
    cin>>n;
    for(int i = 0;i < n;i++) pile[i].push_back(i);
    string s1,s2;
    int a,b;
    while(cin>>s1){
        if(s1 == "quit") break;
        cin>>a>>s2>>b;
        int pa,ha,pb,hb;
        find_block(a,pa,ha);
        find_block(b,pb,hb);
        if(pa == pb) continue;
        else if(s1 == "move" && s2 == "onto") {clear_block(a,pa,ha);clear_block(b,pb,hb);}
        else if(s1 == "move" && s2 == "over") {clear_block(a,pa,ha);}
        else if(s1 == "pile" && s2 == "onto") {clear_block(b,pb,hb);}

        pile_on(a,pa,ha,pb);
    }
    print();
    //End;
    //Print;
    return 0;
}