hdu 3309
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Roll The Cube
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 91 Accepted Submission(s): 40
Problem Description
This is a simple game.The goal of the game is to roll two balls to two holes each.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up.
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up.
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.
Input
First there's an integer T(T<=100) indicating the case number.
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).
Output
The minimum times you press to achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.
Sample Input
46 3****B**B**H**H****4 4*****BB**HH*****4 4*****BH**HB*****5 6*******.BB***.H*H**..*.*******
Sample Output
312Sorry , sir , my poor program fails to get an answer.
BFS，与以往不同的是，这个题目中要求两个目标从不同的起点到达终点。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAXN 22
typedef unsigned char que[MAXN * MAXN * MAXN * MAXN * 4];
const int fx[] = {-1, 1, 0, 0};
const int fy[] = { 0, 0,-1, 1};
char mat[MAXN][MAXN];
bool f[MAXN][MAXN][MAXN][MAXN][2][2]; // 注意，入洞的状态需要加维度
que qx1, qx2, qy1, qy2, step, ql1, ql2;
int sx1, sy1, sx2, sy2;
int s, e, w, h;
void bfs() {
s = 0; e = 1;
ql1[0] = 0; ql2[0] = 0;
qx1[0] = sx1; qy1[0] = sy1; qx2[0] = sx2; qy2[0] = sy2; step[0] = 0;
f[sx1][sy1][sx2][sy2][0][0] = true;
int nx1, ny1, nx2, ny2, tx1, ty1, tx2, ty2, l1, l2;
while (s != e) {
nx1 = qx1[s]; ny1 = qy1[s]; nx2 = qx2[s]; ny2 = qy2[s];
if (ql1[s])
for (int i = 0; i < 4; ++i) {
tx2 = nx2 + fx[i]; ty2 = ny2 + fy[i];
switch (mat[tx2][ty2]) {
case 'H':
if (tx2 != nx1 || ty2 != ny1) {
printf("%d/n", step[s] + 1);
return;
} else {
if (!f[nx1][ny1][tx2][ty2][1][0]) {
qx1[e] = nx1; qy1[e] = ny1;
qx2[e] = tx2; qy2[e] = ty2;
ql1[e] = 1; ql2[e] = 0;
step[e] = step[s] + 1;
++e;
f[nx1][ny1][tx2][ty2][1][0] = true;
}
}
case '.': case'B': // 注意，B 也是能走的
if (!f[nx1][ny1][tx2][ty2][1][0]) {
qx1[e] = nx1; qy1[e] = ny1;
qx2[e] = tx2; qy2[e] = ty2;
ql1[e] = 1; ql2[e] = 0;
step[e] = step[s] + 1;
++e;
f[nx1][ny1][tx2][ty2][1][0] = true;
}
}
}
else if (ql2[s])
for (int i = 0; i < 4; ++i) {
tx1 = nx1 + fx[i]; ty1 = ny1 + fy[i];
switch (mat[tx1][ty1]) {
case 'H':
if (tx1 != nx2 || ty1 != ny2) {
printf("%d/n", step[s] + 1);
return;
} else {
if (!f[tx1][ty1][nx2][ny2][0][1]) {
qx1[e] = tx1; qy1[e] = ty1;
qx2[e] = nx2; qy2[e] = ny2;
ql1[e] = 0; ql2[e] = 1;
step[e] = step[s] + 1;
++e;
f[tx1][ty1][nx2][ny2][0][1] = true;
}
}
case '.': case 'B': // 同上
if (!f[tx1][ty1][nx2][ny2][0][1]) {
qx1[e] = tx1; qy1[e] = ty1;
qx2[e] = nx2; qy2[e] = ny2;
ql1[e] = 0; ql2[e] = 1;
step[e] = step[s] + 1;
++e;
f[tx1][ty1][nx2][ny2][0][1] = true;
}
}
}
else
for (int i = 0; i < 4; ++i) {
tx1 = nx1 + fx[i]; ty1 = ny1 + fy[i]; tx2 = nx2 + fx[i]; ty2 = ny2 + fy[i];
if (mat[tx1][ty1] == '*') {
tx1 = nx1; ty1 = ny1;
}
if (mat[tx2][ty2] == '*') {
tx2 = nx2; ty2 = ny2;
}
if (tx1 != tx2 || ty1 != ty2) {
if (mat[tx1][ty1] == 'H' && mat[tx2][ty2] == 'H') {
printf("%d/n", step[s] + 1);
return;
} else {
l1 = 0; l2 = 0;
if (mat[tx1][ty1] == 'H')
l1 = 1;
if (mat[tx2][ty2] == 'H')
l2 = 1;
if (!f[tx1][ty1][tx2][ty2][l1][l2]) {
qx1[e] = tx1; qy1[e] = ty1;
qx2[e] = tx2; qy2[e] = ty2;
ql1[e] = l1; ql2[e] = l2;
step[e] = step[s] + 1;
++e;
f[tx1][ty1][tx2][ty2][l1][l2] = true;
}
}
}
}
++s;
}
puts("Sorry , sir , my poor program fails to get an answer.");
}
int main() {
int t; char th;
scanf("%d", &t);
for (int cas = 1; cas <= t; ++cas) {
sx1 = 0;
memset(f, 0, sizeof(f));
scanf("%d%d", &h, &w);
for (int i = 0; i < h; ++i) {
scanf("%c", &th);
for (int j = 0; j < w; ++j) {
scanf("%c", &mat[i][j]);
if (mat[i][j] == 'B') {
if (sx1 == 0) {
sx1 = i; sy1 = j;
} else {
sx2 = i; sy2 = j;
}
}
}
}
bfs();
}
return 0;
}
