POJ 1077 Eight（DFS + IDA*）
题意：
一个3*3的棋盘里，给8个格子填上1-8这8个数，另一个格子空着。每一步，可以将与空格相邻的格子与空格换位，换位方式可间接体现成空格进行上下左右的移动。先给定一个棋盘状态，问能否经过若干步后，将棋盘变成第一行123，第二行456，第三行78空的状态。若能，输出任意一种可行步骤。
思路：
1. IDA* ： f(s) = h(s) + g(s); 可以理解 f 为在状态 s 时距离目标状态的理想距离，f 单调增，并且 f 越小越有可能是我们想要的;
2. 初始状态的 bound 为初始状态到目标状态曼哈顿距离，newbound 为每次 DFS 之后所能达到的理想状态，当然 newbound 越小越好;
3. bound = newbound，继续递归，知道 succ = true 为止;
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 362880 + 10;
const int INFS = 0x7fffffff;
const int fac[15] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
const char dir[5] = {0, 'u', 'd', 'l', 'r'};
const int dis[9][2] = {{0,0}, {0,1}, {0,2}, {1,0}, {1,1}, {1,2}, {2,0}, {2,1}, {2,2}};
bool vis[MAXN], succ;
int Q[MAXN], QC, ENDS;
struct ST {
char e[10];
};
int getvalue(const char e[]) {
int ans = 0, count;
for (int i = 0; i < 8; i++) {
count = 0;
for (int j = i + 1; j < 9; j++) {
if (e[i] > e[j]) count += 1;
}
ans += fac[8-i] * count;
}
return ans;
}
bool challenge(ST& s, int pos, int op) {
if (op == 1) {
// up
if (pos >= 3) {
swap(s.e[pos-3], s.e[pos]);
return true;
}
} else if (op == 2) {
// down
if (pos < 6) {
swap(s.e[pos], s.e[pos+3]);
return true;
}
} else if (op == 3) {
// left
if (pos % 3) {
swap(s.e[pos-1], s.e[pos]);
return true;
}
} else if (op == 4) {
// right
if ((pos + 1) % 3) {
swap(s.e[pos], s.e[pos+1]);
return true;
}
}
return false;
}
int getdiff(const char e[]) {
int ret = 0;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
int k = 3 * i + j;
if (e[k] != 9) {
ret += abs(i - dis[e[k]-1][0]) + abs(j - dis[e[k]-1][1]);
}
}
}
return ret;
}
int dfs(ST& u, int g, int bound) {
int h = getdiff(u.e);
if (h == 0 || succ) {
succ = true;
return g;
}
if (g + h > bound) {
return g + h;
}
int newbound = INFS;
int pos;
for (int i = 0; i < 9; i++)
if (u.e[i] == 9) pos = i;
for (int i = 1; i <= 4; i++) {
ST v = u;
if (challenge(v, pos, i)) {
int state = getvalue(v.e);
if (!vis[state]) {
vis[state] = true;
Q[QC++] = i;
int b = dfs(v, g + 1, bound);
if (succ)
return b;
QC -= 1;
newbound = min(b, newbound);
vis[state] = false;
challenge(v, pos, i);
}
}
}
return newbound;
}
int main() {
ST u, E;
for (int i = 0; i < 9; i++)
E.e[i] = i + 1;
ENDS = getvalue(E.e);
char ch[100];
while (gets(ch)) {
for (int i = 0, j = 0; ch[i]; i++) {
if (ch[i] != ' ') {
if (ch[i] == 'x')
u.e[j++] = 9;
else
u.e[j++] = ch[i] - '0';
}
}
if (getvalue(u.e) == ENDS) {
printf("\n");
continue;
}
succ = false;
int bound = getdiff(u.e);
while (!succ) {
QC = 0;
vis[getvalue(u.e)] = true;
bound = dfs(u, 0, bound);
vis[getvalue(u.e)] = false;
}
for (int i = 0; i < QC; i++)
printf("%c", dir[Q[i]]);
printf("\n");
}
return 0;
}