AOJ 0121 Seven Puzzle （bfs）

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题意：

给你8个格子 0那个格子可以上下左右的动 - - 一开始以为所有格子都能动毕竟日文题目那么就很像8数码问题了

由其他格子到“01234567” - - 如果挨个bfs的话 case多肯定TLE 那么思维反向一下由“01234567”到其他状态把每个最小步数存到map中对于每个case直接查找就好了

AC代码如下：

////  AOJ 0121 Seven Puzzle////  Created by TaoSama on 2015-02-20//  Copyright (c) 2014 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#define CLR(x,y) memset(x, y, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;string s;typedef pair<string, int> Sta;map<string, int> dp;int d[4] = { -1, 1, 4, -4};void bfs() {queue<Sta> q;dp["01234567"] = 0; q.push(Sta("01234567", 0));while(!q.empty()) {string s = q.front().first;int cur = q.front().second;q.pop();for(int i = 0; i < 4; ++i) {string ns = s;int nxt = cur + d[i];if(nxt < 0 || nxt > 7 || cur == 3 && nxt == 4 || cur == 4 && nxt == 3)continue;swap(ns[cur], ns[nxt]);if(!dp.count(ns)) {dp[ns] = dp[s] + 1;q.push(Sta(ns, nxt));}}}}int main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_base::sync_with_stdio(0);bfs();while(cin >> s) {for(int i = 1; i <= 7; ++i) {char x; cin >> x;s += x;}cout << dp[s] << endl;}return 0;}

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