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  • ZOJ 3814 Sawtooth Puzzle BFS

    感觉可以用BFS撸,然后就撸了,样例无限不过,代码能力真是弱。。

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <set>
    #include <queue>
    
    using namespace std;
    
    const int bufsize = 128;
    char buf[bufsize][bufsize];
    
    struct Block {
    	char str[10][10];
    	void Rotate() {
    		for (int i = 0; i < 8; i++) {
    			for (int j = i + 1; j < 8; j++) {
    				swap(str[i][j], str[j][i]);
    			}
    		}
    		for (int i = 0; i < 8; i++) {
    			for (int j = 0; j < 4; j++) {
    				swap(str[i][j], str[i][8 - j - 1]);
    			}
    		}
    	}
    
    	bool operator == (const Block &x) const {
    		for (int i = 0; i < 8; i++) {
    			for (int j = 0; j < 8; j++) {
    				if (str[i][j] != x.str[i][j]) {
    					return false;
    				}
    			}
    		}
    		return true;
    	}
    };
    
    void getBlock(Block block[9]) {
    	for (int i = 0; i < 24; i++) {
    		for (int j = 0; j < 24; j++) {
    			scanf(" %c", &buf[i][j]);
    		}
    	}
    	int id = 0;
    	for (int i = 0; i < 24; i += 8) {
    		for (int j = 0; j < 24; j += 8) {
    			for (int dx = 0; dx < 8; dx++) {
    				for (int dy = 0; dy < 8; dy++) {
    					block[id].str[dx][dy] = buf[i + dx][j + dy];
    				}
    			}
    			id++;
    		}
    	}
    }
    
    bool finalst[1 << 18];
    Block ori[9], tar[9];
    int e[9];
    
    void make_hash(int pos, int val) {
    	if (pos == -1) {
    		finalst[val] = true;
    		return;
    	}
    	for (int i = 0; i < 4; i++) {
    		if (ori[pos] == tar[pos]) {
    			make_hash(pos - 1, val << 2 | i);
    		}
    		ori[pos].Rotate();
    	}
    }
    
    void input() {
    	getBlock(ori); getBlock(tar);
    	memset(e, 0, sizeof(e));
    	for (int i = 0; i < 9; i++) {
    		for (int j = 0; j < 4; j++) {
    			int tmp; scanf("%d", &tmp);
    			e[i] |= (tmp << j);
    		}
    	}
    }
    
    bool vis1[9];
    const int dx[4] = { 0, -1, 0, 1 };
    const int dy[4] = { -1, 0, 1, 0 };
    
    int gete(int state, int pos) {
    	int mask = 3 << (pos << 1), val = (state & mask) >> (pos << 1);
    	int ret = e[pos];
    	for (int i = 0; i < val; i++) {
    		int bit = (ret & (1 << 3)) >> 3;
    		ret &= (1 << 3) - 1;
    		ret <<= 1; ret |= bit;
    	}
    	return ret;
    }
    
    void dfs(int &nstate, int state, int pos, int d) {
    	vis1[pos] = true;
    
    	int mask = 3 << (pos << 1), val = (state & mask) >> (pos << 1);
    	if (d == 0) val = (val + 1) % 4;
    	else val = (val + 3) % 4;
    	nstate &= ~mask;  nstate |= val << (pos << 1);
    
    	int x = pos / 3, y = pos % 3;
    	for (int i = 0; i < 4; i++) {
    		int nx = x + dx[i], ny = y + dy[i], npos = nx * 3 + ny;
    		if (nx < 0 || nx > 2 || ny < 0 || ny > 2) continue;
    		if (vis1[npos]) continue;
    		int e1 = gete(state, pos), e2 = gete(state, npos);
    		if ((e1 & (1 << i)) && (e2 & (1 << ((i + 2) % 4)))) {
    			dfs(nstate, state, npos, d ^ 1);
    		}
    	}
    }
    
    int Rotate(int state, int pos) {
    	memset(vis1, 0, sizeof(vis1));
    	int ret = state;
    	dfs(ret, state, pos, 0);
    	return ret;
    }
    
    bool vis[1 << 18];
    
    void solve() {
    	memset(finalst, 0, sizeof(finalst));
    	memset(vis, 0, sizeof(vis));
    	make_hash(8, 0);
    	queue<int> q, qd;
    	q.push(0); qd.push(0);
    	vis[0] = true;
    	while (!q.empty()) {
    		int nowstate = q.front(); q.pop();
    		int nowdist = qd.front(); qd.pop();;
    		if (finalst[nowstate]) {
    			printf("%d
    ", nowdist);
    			return;
    		}
    		for (int i = 0; i < 9; i++) {
    			int nstate = Rotate(nowstate, i);
    			if (!vis[nstate]) {
    				q.push(nstate);
    				qd.push(nowdist + 1);
    				vis[nstate] = true;
    			}
    		}
    	}
    	puts("-1");
    }
    
    int main() {
    	int T; scanf("%d", &T);
    	while (T--) {
    		input();
    		solve();
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/3960726.html
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