POJ 1077 Eight（八数码第八境界|IDA*+曼哈顿距离+判断是否有解）

来源：互联网发布：淘宝搜索的算法编辑：程序博客网时间：2024/05/21 19:35

题意：八数码。

思路：IDA*+曼哈顿距离+判断是否有解。因为要用到IDA*搜索，所以我们搜索之前先判断一下是否有解。

判断的方法是学习一个大神的：判断八数码问题是否有解

IDA*比起BFS的好处是空间复杂度极低，同时因为有剪枝，比起BFS降低了盲目性；比起A*的好处是不用去维护一个堆。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<map>#include<set>#include<ctime>#define eps 1e-6#define LL long long#define pii pair<int, int>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 400000;//const int INF = 0x3f3f3f3f;int dx[] = {-1, 0, 1, 0};int dy[] = {0, -1, 0, 1};int mov[1000];int maxd;char c;struct State {int s[9];int pos;} state;int cal_pos(int pos, int i) {int nx = pos/3+dx[i], ny = pos%3+dy[i];if(nx<0 || nx>2 || ny<0 || ny>2) return -1;return nx*3 + ny;}int cal_h() {int ans = 0;for(int i = 0; i < 9; i++) {if(state.s[i] == 9) continue;int x = (state.s[i]-1)/3, y = (state.s[i]-1)%3;int nx = i/3, ny = i%3;ans += abs(x-nx) + abs(y-ny);}return ans;}void init() {if(c != 'x') state.s[0] = c - '0';else {state.s[0] = 9;state.pos = 0;}for(int i = 1; i < 9; i++) {cin >> c;if(c == 'x') {state.s[i] = 9;state.pos = i;}else state.s[i] = c - '0';}}bool dfs(int d, int last) {if(d == maxd) {if(!cal_h()) return true;return false;}for(int i = 0; i < 4; i++) {if(abs(i-last) == 2) continue;int newpos = cal_pos(state.pos, i);if(newpos<0) continue;swap(state.s[state.pos], state.s[newpos]);swap(state.pos, newpos);if(d+cal_h() > maxd) {swap(state.s[state.pos], state.s[newpos]);swap(state.pos, newpos);continue;}mov[d+1] = i;if(dfs(d+1, i)) return true;swap(state.s[state.pos], state.s[newpos]);swap(state.pos, newpos);}return false;} void IDA_star() {for(maxd = 0; ; maxd++) {if(dfs(0, 100)) return;}}void print_path(int dep) {if(!dep) return;print_path(dep-1);if(!mov[dep]) printf("u");else if(mov[dep]==1) printf("l");else if(mov[dep]==2) printf("d");else printf("r");}bool is_solvable() {int cnt = 0;for(int i = 0; i < 9; i++) {if(state.s[i] == 9) continue;for(int j = 0; j < i; j++) {if(state.s[j] == 9) continue;if(state.s[j] > state.s[i]) cnt++;}}return !(cnt%2);}int main() {    //freopen("input.txt", "r", stdin);while(cin >> c) { init();if(!is_solvable()) printf("unsolvable");else {IDA_star();print_path(maxd);}puts("");}    return 0;}

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