29 | 12月 | 2017 | 超客的水墨烟雨

uva101

maksyuki 发表于 oj 分类，标签: 基础算法-模拟

29 12月 2017

Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies. For example, an early AI study of planning and robotics (STRIPS) used a block world in which a robot arm performed tasks involving the manipulation of blocks.

In this problem you will model a simple block world under certain rules and constraints. Rather than determine how to achieve a specified state, you will “program” a robotic arm to respond to a limited set of commands.

The problem is to parse a series of commands that instruct a robot arm in how to manipulate blocks that lie on a flat table. Initially there are n blocks on the table (numbered from 0 to n − 1) with block bi adjacent to block bi+1 for all 0 ≤ i < n − 1 as shown in the diagram below:

The valid commands for the robot arm that manipulates blocks are:

• move a onto b

where a and b are block numbers, puts block a onto block b after returning any blocks that are stacked on top of blocks a and b to their initial positions.

• move a over b

where a and b are block numbers, puts block a onto the top of the stack containing block b, after returning any blocks that are stacked on top of block a to their initial positions.

• pile a onto b

where a and b are block numbers, moves the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto block b. All blocks on top of block b are moved to their initial positions prior to the pile taking place. The blocks stacked above block a retain their order when moved.

• pile a over b

where a and b are block numbers, puts the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto the top of the stack containing block b. The blocks stacked above block a retain their original order when moved.

• quit

terminates manipulations in the block world.

Any command in which a = b or in which a and b are in the same stack of blocks is an illegal command. All illegal commands should be ignored and should have no affect on the configuration of blocks.

Input

The input begins with an integer n on a line by itself representing the number of blocks in the block world. You may assume that 0 < n < 25.

The number of blocks is followed by a sequence of block commands, one command per line. Your program should process all commands until the quit command is encountered.

You may assume that all commands will be of the form specified above. There will be no syntactically incorrect commands.

Output

The output should consist of the final state of the blocks world. Each original block position numbered i (0 ≤ i < n where n is the number of blocks) should appear followed immediately by a colon. If there is at least a block on it, the colon must be followed by one space, followed by a list of blocks that appear stacked in that position with each block number separated from other block numbers by a space. Don’t put any trailing spaces on a line.

There should be one line of output for each block position (i.e., n lines of output where n is the integer on the first line of input).

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
Sample Output

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

题目类型：简单模拟

算法分析：用不定长数组进行计算，本题直接进行模拟即可。合理设计函数可以减少重复编码

/**************************************************
filename       :g.cpp
author         :maksyuki
created time   :2017/12/29 22:48:03
last modified  :2017/12/29 23:25:00
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 66;

vector<int> p[maxn];
int n;

int findp(int v) {
    for(int i = 0; i < n; i++) {
        int len = p[i].size();
        for(int j = 0; j < len; j++) 
            if(v == p[i][j]) return i;
    }
    return -1;
}

void clearp(int va, int vb) {
    int len = p[va].size();
    int pos = -1;
    for(int i = 0; i < len; i++)
        if(p[va][i] == vb) {
            pos = i; break;
        }

    for(int i = pos + 1; i < len; i++) p[p[va][i]].push_back(p[va][i]);
    p[va].resize(pos + 1);
}

void putp(int va, int vva, int vb) {
    int pa = -1, len = p[va].size();
    for(int i = 0; i < len; i++)
        if(p[va][i] == vva) {
            pa = i; break;
        }
    
    for(int i = pa; i < len; i++) p[vb].push_back(p[va][i]);
    p[va].resize(pa);
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    while(scanf("%d", &n) != EOF) {
        char op1[16], op2[16];
        int va, vb;
        for(int i = 0; i < maxn; i++) p[i].clear();
        for(int i = 0; i < n; i++) p[i].push_back(i);
        
        while(scanf(" %s", op1)) {
            if(!strcmp(op1, "quit")) break;

            scanf(" %d %s %d", &va, op2, &vb);
            int pa = findp(va);
            int pb = findp(vb);
            if(pa == pb) continue;

            if(!strcmp(op1, "move")) clearp(pa, va);
            if(!strcmp(op2, "onto")) clearp(pb, vb);

            putp(pa, va, pb);
        }
        for(int i = 0; i < n; i++) {
            printf("%d:", i);
            int len = p[i].size();
            for(int j = 0; j < len; j++) printf(" %d", p[i][j]);
            puts("");
        }
    }
    return 0;
}

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/**************************************************

filename :g.cpp

author :maksyuki

created time :2017/12/29 22:48:03

last modified :2017/12/29 23:25:00

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 66;

vector<int> p[maxn];

int n;

int findp(int v) {

for(int i = 0; i < n; i++) {

int len = p[i].size();

for(int j = 0; j < len; j++)

if(v == p[i][j]) return i;

}

return -1;

}

void clearp(int va, int vb) {

int len = p[va].size();

int pos = -1;

for(int i = 0; i < len; i++)

if(p[va][i] == vb) {

pos = i; break;

}

for(int i = pos + 1; i < len; i++) p[p[va][i]].push_back(p[va][i]);

p[va].resize(pos + 1);

}

void putp(int va, int vva, int vb) {

int pa = -1, len = p[va].size();

for(int i = 0; i < len; i++)

if(p[va][i] == vva) {

pa = i; break;

}

for(int i = pa; i < len; i++) p[vb].push_back(p[va][i]);

p[va].resize(pa);

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

while(scanf("%d", &n) != EOF) {

char op1[16], op2[16];

int va, vb;

for(int i = 0; i < maxn; i++) p[i].clear();

for(int i = 0; i < n; i++) p[i].push_back(i);

while(scanf(" %s", op1)) {

if(!strcmp(op1, "quit")) break;

scanf(" %d %s %d", &va, op2, &vb);

int pa = findp(va);

int pb = findp(vb);

if(pa == pb) continue;

if(!strcmp(op1, "move")) clearp(pa, va);

if(!strcmp(op2, "onto")) clearp(pb, vb);

putp(pa, va, pb);

}

for(int i = 0; i < n; i++) {

printf("%d:", i);

int len = p[i].size();

for(int j = 0; j < len; j++) printf(" %d", p[i][j]);

puts("");

}

return 0;

}

uva10474

maksyuki 发表于 oj 分类，标签: 基础算法-排序

29 12月 2017

Raju and Meena love to play with Marbles. They have got a lot of marbles with numbers written on them. At the beginning, Raju would place the marbles one after another in ascending order of the numbers written on them. Then Meena would ask Raju to find the first marble with a certain number. She would count 1...2...3. Raju gets one point for correct answer, and Meena gets the point if Raju fails. After some fixed number of trials the game ends and the player with maximum points wins. Today it’s your chance to play as Raju. Being the smart kid, you’d be taking the favor of a computer. But don’t underestimate Meena, she had written a program to keep track how much time you’re taking to give all the answers. So now you have to write a program, which will help you in your role as Raju.

Input

There can be multiple test cases. Total no of test cases is less than 65. Each test case consists begins with 2 integers: N the number of marbles and Q the number of queries Mina would make. The next N lines would contain the numbers written on the N marbles. These marble numbers will not come in any particular order. Following Q lines will have Q queries. Be assured, none of the input numbers are greater than 10000 and none of them are negative.

Input is terminated by a test case where N = 0 and Q = 0.

Output

For each test case output the serial number of the case.

For each of the queries, print one line of output. The format of this line will depend upon whether or not the query number is written upon any of the marbles. The two different formats are described below:

• ‘x found at y’, if the first marble with number x was found at position y. Positions are numbered 1, 2, . . . , N.

• ‘x not found’, if the marble with number x is not present. Look at the output for sample input for details.

Sample Input

4 1
2
3
5
1
5
5 2
1
3
3
3
1
2
3
0 0

Sample Output

CASE# 1:

5 found at 4

CASE# 2:

2 not found

3 found at 3

题目类型：简单排序

算法分析：将数据按照递增序排好后使用二分查找是否存在给定值即可

/**************************************************
filename       :f.cpp
author         :maksyuki
created time   :2017/12/29 21:48:31
last modified  :2017/12/29 21:57:20
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

int aa[maxn];

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    int N, Q, id = 1;
    while(scanf("%d %d", &N, &Q) != EOF) {
        if(!N && !Q) break;

        for(int i = 1; i <= N; i++) scanf(" %d", &aa[i]);
        sort(aa + 1, aa + 1 + N);
//1 2 3 5    
        int tmp;
        printf("CASE# %d:\n", id++);
        for(int i = 1; i <= Q; i++) {
            scanf(" %d", &tmp);
            int p = lower_bound(aa + 1, aa + 1 + N, tmp) - (aa + 1);
            if(aa[p+1] == tmp) printf("%d found at %d\n", tmp, p + 1);
            else printf("%d not found\n", tmp);
        }
    }
    return 0;
}

/**************************************************

filename :f.cpp

author :maksyuki

created time :2017/12/29 21:48:31

last modified :2017/12/29 21:57:20

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

int aa[maxn];

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int N, Q, id = 1;

while(scanf("%d %d", &N, &Q) != EOF) {

if(!N && !Q) break;

for(int i = 1; i <= N; i++) scanf(" %d", &aa[i]);

sort(aa + 1, aa + 1 + N);

//1 2 3 5

int tmp;

printf("CASE# %d:\n", id++);

for(int i = 1; i <= Q; i++) {

scanf(" %d", &tmp);

int p = lower_bound(aa + 1, aa + 1 + N, tmp) - (aa + 1);

if(aa[p+1] == tmp) printf("%d found at %d\n", tmp, p + 1);

else printf("%d not found\n", tmp);

}

return 0;

}

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