P1379 八数码问题

aoapc上的八数码问题,在luogu上也有类似的题,p1379,经典题目,lrj给出了一个算法,同时给出了三种判重的方法。本来想用std::queue改写一下,但是出了各种问题,只好抄代码ac掉这道题了。。。

#include <bits/stdc++.h>
using namespace std;
typedef int State[9];
const int maxstate = 1000000;
State st[maxstate], goal;
int dist[maxstate];
set<int> vis;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
void init_lookup_table() {vis.clear();}
int try_to_insert(int s) {
	int v = 0;
	for(int i = 0; i < 9; i++) v = v*10+st[s][i];
	if(vis.count(v)) return 0;
	vis.insert(v);
	return 1;
}
int bfs() {
	init_lookup_table();
	int front = 1, rear = 2;
	while(front < rear) {
		State& s = st[front];
		if(memcmp(goal, s, sizeof(s)) == 0) return front;
		int z;
		for(z = 0; z < 9; z++) if(!s[z]) break;
		int x = z/3;
		int y = z%3;
		for(int d = 0; d < 4; d++) {
			int newx = x + dx[d];
			int newy = y + dy[d];
			int newz = newx * 3 + newy;
			if(newx >= 0 && newy < 3 && newy >= 0 && newy < 3) {
				State& t = st[rear];
				memcpy(&t, &s, sizeof(s));
				t[newz] = 0;
				t[z] = s[newz];
				dist[rear] = dist[front] + 1;
				if(try_to_insert(rear)) rear++;
			}
		}
		front++;
	}
	return 0;
}
int main() {
//	for(int i = 0; i < 9; i++) scanf("%d", &st[1][i]);
	st[1][0] = 2;st[1][1] = 8;st[1][2] = 3;st[1][3] = 1;
	st[1][4] = 9;st[1][5] = 4;st[1][6] = 7;st[1][7] = 6;
	st[1][8] = 5; 
	for(int i = 0; i < 9; i++)	scanf("%d", &goal[i]);
	int ans = bfs();
	if(ans > 0) cout << dist[ans];
	else cout << -1 << endl;
	return 0;
}

posted on 2016-10-29 08:34  蒟蒻konjac  阅读(72)  评论(0编辑  收藏

统计