「NOIP2011」Mayan游戏

传送门
Luogu

解题思路

爆搜，并考虑几个剪枝。

• 不交换颜色相同的方块（有争议，但是可以过联赛数据 \(Q \omega Q\)）
• 左边为空才往左换
• 右边不为空才往右换

因为对于两个相邻方块，右边往左换和左边往右换是等价的，同时还可以保证字典序最小。

细节注意事项

• 爆搜题，你们懂的。。。

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while (!isdigit(c)) f |= c == '-', c = getchar();
	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
	s = f ? -s : s;
}

int n, a[10][10], la[10][10][10], ans[10][10], mark[10][10];

inline bool remv() {
	int flag = 0;
	for (rg int i = 1; i <= 5; ++i)
		for (rg int j = 1; j <= 7; ++j)
			if (a[i][j]) {
				if (i > 1 && i < 5 && a[i][j] == a[i - 1][j] && a[i][j] == a[i + 1][j])
					flag = 1, mark[i][j] = mark[i - 1][j] = mark[i + 1][j] = 1;
				if (j > 1 && j < 7 && a[i][j] == a[i][j - 1] && a[i][j] == a[i][j + 1])
					flag = 1, mark[i][j] = mark[i][j - 1] = mark[i][j + 1] = 1;
			}
	if (!flag) return 0;
	for (rg int i = 1; i <= 5; ++i)
		for (rg int j = 1; j <= 7; ++j)
			if (mark[i][j])
				mark[i][j] = 0, a[i][j] = 0;
	return 1;
}

inline void upt() {
	for (rg int i = 1; i <= 5; ++i) {
		int s = 0;
		for (rg int j = 1; j <= 7; ++j) {
			if (!a[i][j]) ++s;
			else if (s) a[i][j - s] = a[i][j], a[i][j] = 0;
		}
	}
}

inline void mv(int i, int j, int t) {
	swap(a[i][j], a[i + t][j]);
	do { upt(); } while (remv());
}

inline bool check() {
	for (rg int i = 1; i <= 5; ++i)
		if (a[i][1]) return 0;
	return 1;
}

inline void dfs(int x) {
	if (check()) {
		for (rg int i = 1; i < x; ++i)
			printf("%d %d %d\n", ans[0][i], ans[1][i], ans[2][i]);
		exit(0);
	}
	if (x > n) return;
	memcpy(la[x], a, sizeof a);
	for (rg int i = 1; i <= 5; ++i)
		for (rg int j = 1; j <= 7; ++j)
			if (a[i][j]) {
				if (i < 5 && a[i + 1][j] != a[i][j]) {
					mv(i, j, 1);
					ans[0][x] = i - 1, ans[1][x] = j - 1, ans[2][x] = 1;
					dfs(x + 1);
					memcpy(a, la[x], sizeof la[x]);
					ans[0][x] = 0, ans[1][x] = 0, ans[2][x] = 0;
				}
				if (i > 1 && a[i - 1][j] == 0) {
					mv(i, j, -1);
					ans[0][x] = i - 1, ans[1][x] = j - 1, ans[2][x] = -1;
					dfs(x + 1);
					memcpy(a, la[x], sizeof la[x]);
					ans[0][x] = 0, ans[1][x] = 0, ans[2][x] = 0;
				}
			}
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n);
	for (rg int i = 1; i <= 5; ++i)
		for (rg int j = 1; j <= 8; ++j) {
			read(a[i][j]); if (a[i][j] == 0) break;
		}
	dfs(1);
	puts("-1");
	return 0;
}

完结撒花 \(qwq\)