算法-程序设计课week2---作业B - Pour Water POJ - 1606

B - Pour Water POJ - 1606

倒水问题 “fill A” 表示倒满A杯，"empty A"表示倒空A杯，“pour A B” 表示把A的水倒到B杯并且把B杯倒满或A倒空。

Input

输入包含多组数据。每组数据输入 A, B, C 数据范围 0 < A <= B 、C <= B <=1000 、A和B互质。

Output

你的程序的输出将由一系列的指令组成。这些输出行将导致任何一个罐子正好包含C单位的水。每组数据的最后一行输出应该是“success”。输出行从第1列开始，不应该有空行或任何尾随空格。

Sample Input

2 7 5
2 7 4

Sample Output

fill B
pour B A
success 
fill A
pour A B
fill A
pour A B
success

Notes

如果你的输出与Sample Output不同，那没关系。对于某个"A B C"本题的答案是多解的，不能通过标准的文本对比来判定你程序的正确与否。 所以本题由 SPJ（Special Judge）程序来判定你写的代码是否正确。

思路与收获
这道题也是广搜，不过情况比上一题多不少，因此代码也长了。倒水总共有六种操作，其中有两个可以细分，总共八种状态。具体状态在代码中有详细注释。

这道题的代码还可以精简一些。

扫描二维码关注公众号，回复： 9777028 查看本文章

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#pragma warning(disable : 4996)
using namespace std;

struct onestep {
  int x;
  int y;  //x,y用来记录杯子中水的状态.跟该元素的xy坐标相对应
  int last_x;
  int last_y;   //last_x，last_y记录路径，也就是前一步操作时杯中水的状态
  int op_type;  //用1，2，3，4，5，6，代表倒水的操作 只有八种状态和六种操作
  //a倒空
  //b倒空
  //a倒满
  //b倒满
  //a入b
  //  a空
  //  b满
  //b入a
  //  a满
  //  b空
  int visited;  //标记
  int num;      //记录截止到这个节点位置倒水多少步
};

void bfs(int cup_a_capacity, int cup_b_capacity, int target_volume) {
  int current_volume_x, current_volume_y;
  /*初始化状态表*/
  onestep** steps = new onestep*[cup_a_capacity + 1];
  for (int i = 0; i <= cup_a_capacity; i++) {
    steps[i] = new onestep[cup_b_capacity + 1];
  }

  for (int i = 0; i <= cup_a_capacity; i++) {
    for (int j = 0; j <= cup_b_capacity; j++) {
      steps[i][j].x = i;
      steps[i][j].y = j;
      steps[i][j].last_x = 0;
      steps[i][j].last_y = 0;
      steps[i][j].op_type = 0;
      steps[i][j].num = 0;
      steps[i][j].visited = 0;
    }
  }

  /*队列，记录走过的广搜路径*/
  queue<onestep> q;
  q.push(steps[0][0]);

  steps[0][0].visited = 1;  //起点打个标记

  while (q.size()) {        //当深搜还有路径可走
    onestep p = q.front();  //取队首
    q.pop();

    current_volume_x = p.x;  //记录当前步骤在状态表中的位置，这也是
    current_volume_y = p.y;
    if (current_volume_x == target_volume || current_volume_y == target_volume) {  //如果当前步骤就是所需结果
      int path[10000];
      memset(path, 0, sizeof(path));
      printf("%d\n", steps[current_volume_x][current_volume_y].num);  //打印步数
      int sum = steps[current_volume_x][current_volume_y].num;

      onestep* current_step = &steps[current_volume_x][current_volume_y];  //记录当前步骤的操作
      for (int i = sum; i > 0; i--) {                                      //从末尾开始整理步骤
        path[i] = current_step->op_type;
        /*获取上一个步骤的操作*/
        current_step = &steps[current_step->last_x][current_step->last_y];
      }
      for (int i = 1; i <= sum; i++) {
        if (path[i] == 1)
          printf("fill A\n");
        else if (path[i] == 2)
          printf("empty A\n");
        else if (path[i] == 3)
          printf("fill B\n");
        else if (path[i] == 4)
          printf("empty B\n");
        else if (path[i] == 5)
          printf("pour A B\n");
        else if (path[i] == 6)
          printf("pour B A\n");
      }
      return;
    }

    int a_left;
    int b_left;
    /*对于当前状态，穷举其接下来的六种操作和八种状态*/
    if (steps[cup_a_capacity][p.y].visited == 0) {  //若当前a杯子不满状态未访问过，则加满杯子a并记录访问
      steps[cup_a_capacity][p.y].last_x = p.x;
      steps[cup_a_capacity][p.y].last_y = p.y;
      steps[cup_a_capacity][p.y].visited = 1;
      steps[cup_a_capacity][p.y].num = steps[p.x][p.y].num + 1;
      steps[cup_a_capacity][p.y].op_type = 1;
      q.push(steps[cup_a_capacity][p.y]);
    }
    if (steps[0][p.y].visited == 0) {  //若杯子a倒空未曾访问，则倒空并记录访问
      steps[0][p.y].last_x = p.x;
      steps[0][p.y].last_y = p.y;
      steps[0][p.y].visited = 1;
      steps[0][p.y].num = steps[p.x][p.y].num + 1;
      steps[0][p.y].op_type = 2;
      q.push(steps[0][p.y]);
    }
    if (steps[p.x][cup_b_capacity].visited == 0) {  //杯子b加满，
      steps[p.x][cup_b_capacity].last_x = p.x;
      steps[p.x][cup_b_capacity].last_y = p.y;
      steps[p.x][cup_b_capacity].visited = 1;
      steps[p.x][cup_b_capacity].num = steps[p.x][p.y].num + 1;
      steps[p.x][cup_b_capacity].op_type = 3;
      q.push(steps[p.x][cup_b_capacity]);
    }
    if (steps[p.x][0].visited == 0) {  //杯子b倒空
      steps[p.x][0].last_x = p.x;
      steps[p.x][0].last_y = p.y;
      steps[p.x][0].visited = 1;
      steps[p.x][0].num = steps[p.x][p.y].num + 1;
      steps[p.x][0].op_type = 4;
      q.push(steps[p.x][0]);
    }

    a_left = p.x + p.y - cup_b_capacity;

    if (a_left >= 0 && steps[a_left][cup_b_capacity].visited == 0) {  //a倒入b，a有剩余
      steps[a_left][cup_b_capacity].last_x = p.x;
      steps[a_left][cup_b_capacity].last_y = p.y;
      steps[a_left][cup_b_capacity].visited = 1;
      steps[a_left][cup_b_capacity].num = steps[p.x][p.y].num + 1;
      steps[a_left][cup_b_capacity].op_type = 5;
      q.push(steps[a_left][cup_b_capacity]);
    }
    if (a_left < 0 && steps[0][p.x + p.y].visited == 0) {  //a倒入b，a空，b不清楚
      steps[0][p.x + p.y].last_x = p.x;
      steps[0][p.x + p.y].last_y = p.y;
      steps[0][p.x + p.y].visited = 1;
      steps[0][p.x + p.y].num = steps[p.x][p.y].num + 1;
      steps[0][p.x + p.y].op_type = 5;
      q.push(steps[0][p.x + p.y]);
    }
    b_left = p.x + p.y - cup_a_capacity;
    if (b_left >= 0 && steps[cup_a_capacity][b_left].visited == 0) {  //b倒入a，a满，b不清楚
      steps[cup_a_capacity][b_left].last_x = p.x;
      steps[cup_a_capacity][b_left].last_y = p.y;
      steps[cup_a_capacity][b_left].visited = 1;
      steps[cup_a_capacity][b_left].num = steps[p.x][p.y].num + 1;
      steps[cup_a_capacity][b_left].op_type = 6;
      q.push(steps[cup_a_capacity][b_left]);
    }
    if (b_left < 0 && steps[p.x + p.y][0].visited == 0) {  //b倒入a，b空，a不清楚
      steps[p.x + p.y][0].last_x = p.x;
      steps[p.x + p.y][0].last_y = p.y;
      steps[p.x + p.y][0].visited = 1;
      steps[p.x + p.y][0].num = steps[p.x][p.y].num + 1;
      steps[p.x + p.y][0].op_type = 6;
      q.push(steps[p.x + p.y][0]);
    }
  }

  //printf("这两个杯子无法称量出%d的水\n", target_volume);
}

int main() {
  int cup_a_capacity, cup_b_capacity, target_volume;
  while (~scanf("%d %d %d", &cup_a_capacity, &cup_b_capacity, &target_volume)) {
    bfs(cup_a_capacity, cup_b_capacity, target_volume);
    printf("success\n");
  }

  return 0;
}