Section 1.4,The Clocks

1题目描述

英文： http://ace.delos.com/usacoprob2?a=2dcMjFnNl7S&S=clocks

中文翻译： http://www.nocow.cn/index.php/Translate:USACO/clocks

2深度优先搜索解法

2.1 思路：用二进制的三位保存一个时钟的状态。低两位表示四种状态，第三位用作进位，每次操作后置零[1][2]

2.2 代码：

/*
ID: wangsiy1
LANG: C
TASK: clocks
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define DEBUG 1

/* 移动方法编号是way的下标加一 */
int way[9] = {
    /*00A00B00C00D00E00F00G00H00I  */
    0b001001000001001000000000000, /* 1 */
    0b001001001000000000000000000, /* 2 */
    0b000001001000001001000000000, /* 3 */
    0b001000000001000000001000000, /* 4 */
    0b000001000001001001000001000, /* 5 */
    0b000000001000000001000000001, /* 6 */
    0b000000000001001000001001000, /* 7 */
    0b000000000000000000001001001, /* 8 */
    0b000000000000001001000001001, /* 9 */
};

int afteradd = 0b00000011011011011011011011011011;

int cantakethis(int *path, int i, int depth)
{
    int m, r=0;
    for(m=0; m<depth; m++) {
    if(path[m]==i)
        ++r;
    }
    if (r>=3) return 0;
    else return 1;
}

void printresult(int *path, int depth)
{
    int i;
    FILE *fout = fopen ("clocks.out", "w");
    for(i=0; i<depth; i++) {
    fprintf(fout, "%d", path[i]);
    if(i==depth-1)
        fprintf(fout, "\n");
    else
        fprintf(fout, " ");
    }
}

/* state初始为0 */
void readdata(FILE *fin, int *state)
{
    int c, i;
    for(i=0; i<9; i++) {
    fscanf(fin, "%d", &c);
    (*state) <<= 3;
    if(c==12)
        c = 0;
    else if (c==3)
        c = 1;
    else if (c==6)
        c = 2;
    else if (c==9)
        c = 3;
    *state |= c;
    }
}


void dfs(int depth, int state, int *path)
{
    int i, firstway, tmpstate;
    if(state==0) {
    printresult(path, depth);
    }
    else {
    if(depth > 0)
        firstway = path[depth-1];
    else
        firstway = 1;
    /* i 方法编号，既下标加一 */
    for(i=firstway; i<=9; i++) {
        if (cantakethis(path, i, depth)) {
        tmpstate = state;
        state += way[i-1];
        state &= afteradd;
        path[depth] = i;
        /* printresult(path, depth); */
        dfs(depth+1, state, path);
        state = tmpstate;
        }
    }
    }
}

main ()
{
    /* path[]记录方法步骤，depth为递归深度,同时为depth的最后一个元素下标 */
    int state=0, path[27], depth=0;
    char s[32];
    FILE *fin  = fopen ("clocks.in", "r");
    

    readdata(fin, &state);
    /* printf("%d\n", state); */
    dfs(0, state, path);
    exit (0);
}

2.3 执行结果

Executing...
   Test 1: TEST OK [0.027 secs, 1972 KB]
   Test 2: TEST OK [0.027 secs, 1972 KB]
   Test 3: TEST OK [0.027 secs, 1972 KB]
   Test 4: TEST OK [0.027 secs, 1972 KB]
   Test 5: TEST OK [0.027 secs, 1972 KB]
   Test 6: TEST OK [0.027 secs, 1972 KB]
   Test 7: TEST OK [0.027 secs, 1972 KB]
   Test 8: TEST OK [0.027 secs, 1972 KB]
   Test 9: TEST OK [0.027 secs, 1972 KB]

3广度优先搜索解法

3.1广搜的队列最大长度为 4^9=262144 [1]

3.2代码

/*
ID: wangsiy1
LANG: C
TASK: clocks
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define DEBUG 0
#define QLEN 262144

typedef struct {
    int state;
    char path[27], len;
} clock;

/* 移动方法编号是way的下标加一 */
int way[9] = {
    /*00A00B00C00D00E00F00G00H00I  */
    0b001001000001001000000000000, /* 1 */
    0b001001001000000000000000000, /* 2 */
    0b000001001000001001000000000, /* 3 */
    0b001000000001000000001000000, /* 4 */
    0b000001000001001001000001000, /* 5 */
    0b000000001000000001000000001, /* 6 */
    0b000000000001001000001001000, /* 7 */
    0b000000000000000000001001001, /* 8 */
    0b000000000000001001000001001, /* 9 */
};

int afteradd = 0b00000011011011011011011011011011;

int cantakethis(clock *t, int i)
{
    int m, r=0, depth = t->len;
    for(m=0; m<depth; m++) {
    if(t->path[m]==i)
        ++r;
    }
    if (r>=3) return 0;
    else return 1;
}

void printresult(clock *t)
{
    int i, len;
    FILE *fout = fopen ("clocks.out", "w");
    len = t->len;
    
    for(i=0; i<len; i++) {
    fprintf(fout, "%hd", t->path[i]);
    if(i==len-1)
        fprintf(fout, "\n");
    else
        fprintf(fout, " ");
    }
}

/* state初始为0 */
void readdata(FILE *fin, int *state)
{
    int c, i;
    for(i=0; i<9; i++) {
    fscanf(fin, "%d", &c);
    (*state) <<= 3;
    if(c==12)
        c = 0;
    else if (c==3)
        c = 1;
    else if (c==6)
        c = 2;
    else if (c==9)
        c = 3;
    *state |= c;
    }
}

void cpyclock(clock *from, clock *to)
{
    int i;
    to->state = from->state;
    to->len = from->len;
    for(i=0; i<from->len; i++)
    to->path[i] = from->path[i];
}

void dequeue(clock *queue, clock *tmp, int *head)
{
    cpyclock(queue+*head, tmp);
    (*head) = (++(*head)) % QLEN;
}

void enqueue(clock *queue, clock *tmp2, int *tail)
{
    cpyclock(tmp2, queue+*tail);
    *tail = (++(*tail)) % QLEN;
}

void bfs(int state)
{
    int head, tail, len, startway, i, j ;
    clock queue[QLEN], *tmp, *tmp2;
    tmp = (clock *) malloc (sizeof(clock));
    tmp2 = (clock *) malloc (sizeof(clock));
    
    queue[0].state=state;
    queue[9].len = 0;
    head = 0;
    tail = 1;
    
    while(head != tail) {
    dequeue(queue, tmp, &head);
    if(tmp->state == 0) {
        printresult(tmp);
        return;
    }
    if(tmp->len>0)
        startway = tmp->path[tmp->len-1];
    else
        startway = 1;
    for(i=startway; i<=9; i++) {
        if (cantakethis(tmp, i)) {
        tmp2->state = tmp->state + way[i-1];
        tmp2->state &= afteradd;
        tmp2->len = tmp->len + 1;
        for(j=0; j<= tmp->len; j++)
            tmp2->path[j] = tmp->path[j];
        tmp2->path[tmp2->len - 1] = i;
#if DEBUG
        /* printf("state: %d", tmp2->state); */
        /* printf("path:"); */
        /* printresult(tmp2); */
        printf("len :%d\n", tmp2->len);
#endif      
        if(tmp2->state == 0) {
            printresult(tmp2);
            return;
        }
        else
            enqueue(queue, tmp2, &tail);
        }
    }
    }
}
main ()
{
    /* times操作次数 */
    int state=0, path[27], times=0;
    char s[32];
    FILE *fin  = fopen ("clocks.in", "r");
    

    readdata(fin, &state);
    /* printf("%d\n", state); */
    bfs(state);
    exit (0);
}

3.3执行结果

Executing...
   Test 1: TEST OK [0.000 secs, 10044 KB]
   Test 2: TEST OK [0.000 secs, 10040 KB]
   Test 3: TEST OK [0.000 secs, 10036 KB]
   Test 4: TEST OK [0.000 secs, 10044 KB]
   Test 5: TEST OK [0.000 secs, 10040 KB]
   Test 6: TEST OK [0.054 secs, 10040 KB]
   Test 7: TEST OK [0.054 secs, 10040 KB]
   Test 8: TEST OK [0.054 secs, 10044 KB]
   Test 9: TEST OK [0.054 secs, 10044 KB]

参考文献：

 [1] http://www.nocow.cn/index.php/USACO/clocks

 [2] http://blog.csdn.net/pennyshe/archive/2011/02/27/6211780.aspx