POJ_3414_BFS pots

/*
* POJ_3414_BFS
* I really like this bfs problem
* On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
*/
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#define BUG puts("here!!!");

using namespace std;
const int N = 205;
struct Node {
int k1, k2;
int steps;
int op;
int pre;
};
bool vis

;
string str[10] = { "FILL(1)", "DROP(1)", "FILL(2)", "DROP(2)", "POUR(1,2)", "POUR(2,1)" };
Node cur, nex;
Node que[N*N];
int A, B, C;
void dfs(Node pn) {
if (pn.pre != 0) {
dfs(que[pn.pre]);
}
cout << str[pn.op] << endl;
}
void bfs(int v1, int v2) {
cur.k1 = v1; cur.k2 = v2; cur.steps = 0; cur.pre = 0;
que[0] = cur;
vis[0][0] = 1;
int head = 0, tail = 1;
while (head < tail) {
Node top = que[head++];
if (top.k1 == C || top.k2 == C) {
cout << top.steps << endl;
if (C != 0) dfs(top);
return;
}
int yi = 0;
for (int i = 0; i < 6; i++) {
switch(i) {
case 0 : nex.k1 = A; nex.k2 = top.k2; break;
case 1 : nex.k1 = 0; nex.k2 = top.k2; break;
case 2 : nex.k1 = top.k1; nex.k2 = B; break;
case 3 : nex.k2 = top.k1; nex.k2 = 0; break;
case 4 :
yi = top.k1 + top.k2 - B;
if(yi > 0) nex.k1 = yi, nex.k2 = B;
else nex.k1 = 0, nex.k2 = B + yi;
break;
case 5 :
yi = top.k1 + top.k2 - A;
if(yi > 0) nex.k1 = A, nex.k2 = yi;
else nex.k1 = A + yi, nex.k2 = 0;
break;
}
nex.op = i;
nex.steps = top.steps + 1;
nex.pre = head - 1;
if (!vis[nex.k1][nex.k2]) {
que[tail++] = nex;
vis[nex.k1][nex.k2] = true;
}
}
}
printf("impossible\n");
}
int main() {
while (scanf("%d%d%d", &A, &B, &C) == 3) {
memset(vis, 0, sizeof(vis));
bfs(0, 0);
}
return 0;
}