HDU 1043 Eight（八数码第五境界|A*+哈希+简单估价函数+打表）

来源：互联网发布：数据导入excel 编辑：程序博客网时间：2024/05/01 05:19
题意：八数码。
思路：八数码第五境界，A*搜索。
#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;bool has[MAXN];int fac[9] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};int dx[] = {-1, 0, 1, 0};int dy[] = {0, -1, 0, 1};int last[MAXN]; short int mov[MAXN];struct State {int f, g;int s[9];int pos;int hash;bool operator < (const State& A) const {return A.f==f ? g>A.g : f>A.f;}} ini, target;int cal_h(int* s1) {int ans = 0;for(int i = 0; i < 9; i++) {if(s1[i] != target.s[i]) ans++;}return ans;}priority_queue<State> q;int Hash(int* s) {int res = 0;for(int i = 0; i < 8; i++) {int cnt = 0;for(int j = i + 1; j < 9; j++) {if(s[i] > s[j]) cnt++; }res += cnt * fac[8-i];}return res;} 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;}void BFS() {for(int i = 0; i < 9; i++) target.s[i] = i+1;target.pos = 8;target.hash = 0;has[0] = 1;q.push(target);while(!q.empty()) {State ha = q.top();q.pop();State tmp;  for(int i = 0; i < 4; i++) {tmp.pos = cal_pos(ha.pos, i);tmp.g = ha.g + 1;if(tmp.pos<0) continue;for(int j = 0; j < 9; j++) {if(j==ha.pos) tmp.s[j] = ha.s[tmp.pos];else if(j == tmp.pos) tmp.s[j] = ha.s[ha.pos];else tmp.s[j] = ha.s[j];}tmp.hash = Hash(tmp.s);if(has[tmp.hash]) continue;tmp.f = ha.f + cal_h(tmp.s);q.push(tmp);has[tmp.hash] = 1;last[tmp.hash] = ha.hash;mov[tmp.hash] = i;}}}void print_path(int x) {if(x==0) return;int i = mov[x];if(!i) printf("d");else if(i==1) printf("r");else if(i==2) printf("u");else printf("l");print_path(last[x]);}int main() {    //freopen("input.txt", "r", stdin);memset(has, 0, sizeof(has));BFS();char tmp;while(cin >> tmp) { if(tmp != 'x') ini.s[0] = tmp - '0';else {ini.s[0] = 9;ini.pos = 0;}for(int i = 1; i < 9; i++) {cin >> tmp;if(tmp == 'x') {ini.s[i] = 9;ini.pos = i;}else ini.s[i] = tmp - '0';}ini.hash = Hash(ini.s);if(!has[ini.hash]) printf("unsolvable");else print_path(ini.hash);puts("");}    return 0;}
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