想不到这是自己的poj的第100题..非常惭愧 现在才163题 又过了一遍专题 和当年不同的代码风格 忘记不可能的情况wa一发..大规模数据慎用string记录路径..

#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<string>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
int a, b, c, k, x, y, z, v[111][111][5];
char s[11][11] = { "","FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)" };
struct viia { int x, y; };
queue<viia>q;
void out(int x, int y)
{
	if (x + y == 0)  return;
	out(v[x][y][1], v[x][y][2]);
	printf("%s\n", s[v[x][y][3]]);
}
int main()
{
	scanf("%d%d%d", &a, &b, &c);
	viia d, e;
	d.x = d.y = 0;
	q.push(d);
	while (!q.empty())
	{
		d = q.front(), q.pop();
		if (d.x != a && !v[a][d.y][0])
		{
			e.x = a, e.y = d.y, q.push(e);
			v[e.x][e.y][0] = v[d.x][d.y][0] + 1;
			v[e.x][e.y][1] = d.x, v[e.x][e.y][2] = d.y;
			v[e.x][e.y][3] = 1;
			if (e.x == c || e.y == c) { x = e.x, y = e.y; z = v[x][y][0]; break; }
		}
		if (d.y != b && !v[d.x][b][0])
		{
			e.x = d.x, e.y = b, q.push(e);
			v[e.x][e.y][0] = v[d.x][d.y][0] + 1;
			v[e.x][e.y][1] = d.x, v[e.x][e.y][2] = d.y;
			v[e.x][e.y][3] = 2;
			if (e.x == c || e.y == c) { x = e.x, y = e.y; z = v[x][y][0]; break; }
		}
		if (d.x && !v[0][d.y][0])
		{
			e.x = 0, e.y = d.y, q.push(e);
			v[e.x][e.y][0] = v[d.x][d.y][0] + 1;
			v[e.x][e.y][1] = d.x, v[e.x][e.y][2] = d.y;
			v[e.x][e.y][3] = 3;
			if (e.x == c || e.y == c) { x = e.x, y = e.y; z = v[x][y][0]; break; }
		}
		if (d.y && !v[d.x][0][0])
		{
			e.x = d.x, e.y = 0, q.push(e);
			v[e.x][e.y][0] = v[d.x][d.y][0] + 1;
			v[e.x][e.y][1] = d.x, v[e.x][e.y][2] = d.y;
			v[e.x][e.y][3] = 4;
			if (e.x == c || e.y == c) { x = e.x, y = e.y; z = v[x][y][0]; break; }
		}
		k = min(d.x, b - d.y);
		if (!v[d.x - k][d.y + k][0])
		{
			e.x = d.x - k, e.y = d.y + k, q.push(e);
			v[e.x][e.y][0] = v[d.x][d.y][0] + 1;
			v[e.x][e.y][1] = d.x, v[e.x][e.y][2] = d.y;
			v[e.x][e.y][3] = 5;
			if (e.x == c || e.y == c) { x = e.x, y = e.y; z = v[x][y][0]; break; }
		}
		k = min(a - d.x, d.y);
		if (!v[d.x + k][d.y - k][0])
		{
			e.x = d.x + k, e.y = d.y - k, q.push(e);
			v[e.x][e.y][0] = v[d.x][d.y][0] + 1;
			v[e.x][e.y][1] = d.x, v[e.x][e.y][2] = d.y;
			v[e.x][e.y][3] = 6;
			if (e.x == c || e.y == c) { x = e.x, y = e.y; z = v[x][y][0]; break; }
		}
	}
	if (z)printf("%d\n", z), out(x, y);
	else puts("impossible");
}