【BZOJ1085】[SCOI2005]骑士精神【搜索】【剪枝】

【题目链接】

大暴搜。

加个剪枝,如果当前棋盘与目标棋盘相差超过限定步数,那么肯定不行了。

注意要保证步数最小,所以要从小到大枚举步数。

/* Footprints In The Blood Soaked Snow */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 7;

int ans[maxn][maxn] = {
	{0, 0, 0, 0, 0, 0},
	{0, 1, 1, 1, 1, 1},
	{0, 0, 1, 1, 1, 1},
	{0, 0, 0, 2, 1, 1},
	{0, 0, 0, 0, 0, 1},
	{0, 0, 0, 0, 0, 0},
};
int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {-1, -2, -2, -1, 1, 2, 2, 1};

int maxs;

inline bool check(int A[maxn][maxn]) {
	for(int i = 1; i <= 5; i++) for(int j = 1; j <= 5; j++)
		if(A[i][j] != ans[i][j]) return 0;
	return 1;
}

inline bool eva(int A[maxn][maxn], int s) {
	int cnt = 0;
	for(int i = 1; i <= 5; i++) for(int j = 1; j <= 5; j++)
		if(A[i][j] != ans[i][j]) {
			cnt++;
			if(cnt + s > maxs) return 0;
		}
	return 1;
}

inline bool dfs(int A[maxn][maxn], int x, int y, int s) {
	if(s == maxs && check(A)) return 1;
	for(int i = 0; i < 8; i++) {
		int xx = x + dx[i], yy = y + dy[i];
		if(xx < 1 || xx > 5 || yy < 1 || yy > 5) continue;
		swap(A[x][y], A[xx][yy]);
		if(eva(A, s) && dfs(A, xx, yy, s + 1)) return 1;
		swap(A[x][y], A[xx][yy]);
	}
	return 0;
}

inline bool solve(int A[maxn][maxn], int x, int y) {
	for(maxs = 1; maxs <= 15; maxs++)
		if(dfs(A, x, y, 0)) return 1;
	return 0;
}	

int main() {
	int T; scanf("%d", &T);
	while(T--) {
		int x, y, A[maxn][maxn];
		for(int i = 1; i <= 5; i++) {
			char str[maxn]; scanf("%s", str + 1);
			for(int j = 1; j <= 5; j++)
				if(str[j] == '*') A[i][j] = 2, x = i, y = j;
				else A[i][j] = str[j] - '0';
		}
		if(solve(A, x, y)) printf("%d\n", maxs);
		else printf("-1\n");
	}
	return 0;
}

一开始写了bfs+hash,结果状态太多,炸了。

/* Footprints In The Blood Soaked Snow */
#include <cstdio>
#include <cstring>
#include <map>
#include <queue>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 7;

int ans[maxn][maxn] = {
	{0, 0, 0, 0, 0, 0},
	{0, 1, 1, 1, 1, 1},
	{0, 0, 1, 1, 1, 1},
	{0, 0, 0, 2, 1, 1},
	{0, 0, 0, 0, 0, 1},
	{0, 0, 0, 0, 0, 0},
};
int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {-1, -2, -2, -1, 1, 2, 2, 1};

struct _data {
	int g[maxn][maxn], x, y, step;
};

LL ed;
map<LL, bool> vis;
queue<_data> q;

inline LL gethash(int u[maxn][maxn]) {
	LL res = 0;
	for(int i = 1; i <= 5; i++) for(int j = 1; j <= 5; j++) {
		res *= 3;
		res += u[i][j];
	}
	return res;
}

inline bool check(int u[maxn][maxn]) {
	int cnt = 0;
	for(int i = 1; i <= 5; i++) for(int j = 1; j <= 5; j++)
		if(u[i][j] != ans[i][j]) {
			cnt++;
			if(cnt > 15) return 0;
		}
	return 1;
}

inline void solve() {
	vis.clear();
	for(; !q.empty(); q.pop());
	_data st; st.step = 0;
	for(int i = 1; i <= 5; i++) {
		char str[maxn]; scanf("%s", str + 1);
		for(int j = 1; j <= 5; j++)
			if(str[j] == '*') st.g[i][j] = 2, st.x = i, st.y = j;
			else st.g[i][j] = str[j] - '0';
	}
	LL QAQ = gethash(st.g);
	if(QAQ == ed) {
		printf("0\n");
		return;
	}
	vis[QAQ] = 1;
	q.push(st);
	
	while(!q.empty()) {
		_data u = q.front(); q.pop();
		if(u.step > 15) {
			printf("-1\n");
			return;
		}
		for(int i = 0; i < 8; i++) {
			int x = u.x + dx[i], y = u.y + dy[i];
			if(x < 1 || x > 5 || y < 1 || y > 5) continue;
			swap(u.g[u.x][u.y], u.g[x][y]);
			LL tmp = gethash(u.g);
			if(!vis[tmp]) {
				if(tmp == ed) {
					printf("%d\n", u.step + 1);
					return;
				}
				vis[tmp] = 1;
				_data t; t = u; t.x = x; t.y = y; t.step++;
				q.push(t);
			}
			swap(u.g[u.x][u.y], u.g[x][y]);
		}
	}
}

int main() {
	int T; scanf("%d", &T);
	ed = gethash(ans);
	while(T--) solve();
	return 0;
}


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