UVa101 - The Blocks Problem

复制代码//UVa101 - The Blocks Problem #include<iostream> #include<cstdio> #include<string> #include<vector> using namespace std; const int maxn = 30; int n; vector<int>pile[maxn]; //找木块a所在的pile和height，以引用的形式返回调用者 void find(int a, int &p, int &h){ for(p = 0; p < n; p++) for(h = 0; h < pile[p].size(); h++) if(pile[p][h] == a) return; } //把p堆高度为h的木块上方的所有木块移回原位 void back(int p, int h){ for(int i = h+1; i < pile[p].size(); i++){ int b = pile[p][i]; pile[b].push_back(b); } pile[p].resize(h+1); } //把p堆高度为h及其上方的木块整体移动到p2的顶部 void doing(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); } int main(){ int a, b; cin >> n; string s1, s2; for(int i = 0; i < n; i++) pile[i].push_back(i); while(cin >> s1 && s1 != "quit"){ cin >> a >> s2 >> b; int pa, pb, ha, hb; find(a, pa, ha); find(b, pb, hb); if(pa == pb) continue; if(s2 == "onto") back(pb, hb); if(s1 == "move") back(pa, ha); doing(pa, ha, pb); } 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"); } return 0; } /* 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 0: 0 1: 1 9 2 4 2: 3: 3 4: 5: 5 8 7 6 6: 7: 8: 9: */

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//UVa101 - The Blocks Problem
#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
using namespace std;

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

//找木块a所在的pile和height，以引用的形式返回调用者
void find(int a, int &p, int &h){
    for(p = 0; p < n; p++)
        for(h = 0; h < pile[p].size(); h++)
            if(pile[p][h] == a) return;
}

//把p堆高度为h的木块上方的所有木块移回原位
void back(int p, int h){
    for(int i = h+1; i < pile[p].size(); i++){
        int b = pile[p][i];
        pile[b].push_back(b);
    }
    pile[p].resize(h+1);
}

//把p堆高度为h及其上方的木块整体移动到p2的顶部
void doing(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);
}

int main(){
    int a, b;
    cin >> n;
    string s1, s2;
    for(int i = 0; i < n; i++) pile[i].push_back(i);
    while(cin >> s1 && s1 != "quit"){
        cin >> a >> s2 >> b;
        int pa, pb, ha, hb; 
        find(a, pa, ha);
        find(b, pb, hb);
        if(pa == pb) continue;
        if(s2 == "onto") back(pb, hb);
        if(s1 == "move") back(pa, ha);
        doing(pa, ha, pb);
    }
    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");
    }
    return 0;
}

/*
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

0: 0
1: 1 9 2 4
2:
3: 3
4:
5: 5 8 7 6
6:
7:
8:
9:
*/