POJ3414 Pots（BFS）

题意：

给两个不同容量的杯子，要求经过最少的操作使其中一个杯子中有cL水

要点：

也就是BFS题，只不过这题每个节点有6种方向

15304559

Seasonal

3414

Accepted

212K

0MS

C++

2024B

2016-03-23 19:34:10

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define min(a,b) a>b?b:a
int a, b, c;
char str[6][100] = { "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)" };
int vis[105][105];//用数组记录是否到过，这样就不会走回头路了，保证输出最短路径
int step;

struct node
{
	int a, b, num, pre;//num用来存储进行的操作
}que[105*105];

void print(int i)//递归倒序输出
{
	if (que[i].pre != -1)//因为达到c后不需要操作，所以没有尾巴需要输出，同理头部也不需要
	{
		print(que[i].pre);
		printf("%s\n", str[que[i].num]);
	}
}
void bfs()
{
	int front = 0, rear = 1;
	que[0].a = que[0].b = 0;
	que[0].pre = -1;
	vis[0][0] = 1;
	while (front < rear)
	{
		int count = rear - front;
		while (count--)//因为要统计操作次数，所以加一个while，其实如果不算操作数不加效果一样的
		{
			node now = que[front];
			if (now.a == c || now.b == c)
			{
				printf("%d\n", step);
				print(front);
				return;
			}
			if (!vis[a][now.b])
			{
				que[rear].num = 0;
				que[rear].pre = front;
				que[rear].a = a;
				que[rear].b = now.b;
				vis[a][now.b] = 1;
				rear++;
			}
			if (!vis[now.a][b])
			{
				que[rear].num = 1;
				que[rear].pre = front;
				que[rear].a = now.a;
				que[rear].b = b;
				vis[now.a][b] = 1;
				rear++;
			}
			if (!vis[0][now.b])
			{
				que[rear].num = 2;
				que[rear].pre = front;
				que[rear].a = 0;
				que[rear].b = now.b;
				vis[0][now.b] = 1;
				rear++;
			}
			if (!vis[now.a][0])
			{
				que[rear].num = 3;
				que[rear].pre = front;
				que[rear].a = now.a;
				que[rear].b = 0;
				vis[now.a][0] = 1;
				rear++;
			}
			int wat1 = min(now.a, b - now.b);//防止溢出，所以两个取小
			if (!vis[now.a - wat1][now.b + wat1])
			{
				que[rear].num = 4;
				que[rear].pre = front;
				que[rear].a = now.a-wat1;
				que[rear].b = now.b+wat1;
				vis[now.a - wat1][now.b + wat1]=1;
				rear++;
			}
			int wat2 = min(a - now.a, now.b);
			if (!vis[now.a + wat2][now.b - wat2])
			{
				que[rear].num = 5;
				que[rear].pre = front;
				que[rear].a = now.a + wat2;
				que[rear].b = now.b - wat2;
				vis[now.a + wat2][now.b - wat2]=1;
				rear++;
			}
			front++;//出队
		}
		step++;
	}
	printf("impossible\n");
}

int main()
{
	while (~scanf("%d%d%d", &a, &b, &c))
	{
		step = 0;
		memset(vis, 0, sizeof(vis));
		bfs();
	}
	return 0;
}