hdu1067 Gap (bfs+hash)

题意：给定一个4*8的矩阵，最左边一列是空的，右边4*7的格子被11~17，21~27，31~37，41~47这28个数字填满。

首先把11，21，31，41移动到最左边一列，然后每次找一个空格，把空格左边加1的数移动到空格的位置。问多少步能走到目标状态。

一看就是bfs，但是状态很大所以要用hash。

每个状态的hash值这样计算：

int getHash(int arr[4][8]) {
long long hash = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 8; j++) {
hash <<= 1;
hash += arr[i][j];
}
}
return hash % 1000007;
}

hash初始为0，遍历数组中的每一个元素，hash先乘一个质数，再加上当前元素的值。最后再模一个大质数。

注意，每次找到空格时，如果左边是17，27，37，47或者是空格就无法移动。

完整代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>

using namespace std;

struct Node{
int map[4][8];
int step;
Node() {
step = 0;
memset(map, 0, sizeof(map));
map[0][0] = 11;
map[1][0] = 21;
map[2][0] = 31;
map[3][0] = 41;
}
};

int T;
Node start;
int goal;

int getHash(int arr[4][8]) {
long long hash = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 8; j++) {
hash <<= 1;
hash += arr[i][j];
}
}
return hash % 1000007;
}

Node find(Node a, int b) {
a.step++;
for (int i = 0; i < 4; i++)
for (int j = 1; j < 8; j++)
if (a.map[i][j] == b) a.map[i][j] = 0;
return a;
}

bool vis[1000011];
int bfs() {
memset(vis, false, sizeof(vis));
int hash = getHash(start.map);
vis[hash] = 1;
queue<Node> q;
q.push(start);
while (!q.empty()) {
Node cur = q.front();
q.pop();
if (getHash(cur.map) == goal) return cur.step;
for (int i = 0; i < 4; i++) {
for (int j = 1; j < 8; j++) {
if (cur.map[i][j] || !(cur.map[i][j - 1] % 10) || cur.map[i][j - 1] % 10 == 7) continue;
Node nex = find(cur, cur.map[i][j - 1] + 1);
nex.map[i][j] = cur.map[i][j - 1] + 1;
hash = getHash(nex.map);
if (!vis[hash]) {
vis[hash] = 1;
q.push(nex);
}
}
}
}
return -1;
}

int main() {
scanf("%d", &T);
int end[4][8] = {{11,12,13,14,15,16,17,0}, {21,22,23,24,25,26,27,0},
{31,32,33,34,35,36,37,0}, {41,42,43,44,45,46,47,0}};
goal = getHash(end);
while (T--) {
start = Node();
for (int i = 0; i < 4; i++) {
for (int j = 1; j <= 7; j++) {
scanf("%d", &start.map[i][j]);
if (start.map[i][j] % 10 == 1) start.map[i][j] = 0;
}
}
int ans = bfs();
cout << ans << endl;
}
return 0;
}