hdu 5374 Tetris（模拟）

题目链接：hdu 5374 Tetris

模拟。每次进行操作时推断操作是否合法，合法才运行，否则跳过。每次一个token落地，推断一下是否有消除整行。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

/******* Token **********/
const int C[3] = {1, 2, 4};
const int T[3][4][4][2] = {
	{{{0, 0}, {0, 1}, {1, 0}, {1, 1}}, {}, {}, {}}, 
	{{{0, 0}, {0, 1}, {0, 2}, {0, 3}}, {{0, 0}, {1, 0}, {2, 0}, {3, 0}}, {}, {}},
	{{{0, 0}, {0, 1}, {1, 0}, {2, 0}}, {{0, 0}, {0, 1}, {0, 2}, {1, 2}}, {{0, 1}, {1, 1}, {2, 0}, {2, 1}}, {{0, 0}, {1, 0}, {1, 1}, {1, 2}}}
};

void put (const int a[4][2]) {
	int g[4][4];
	memset(g, 0, sizeof(g));

	for (int i = 0; i < 4; i++)
		g[a[i][0]][a[i][1]] = 1;

	for (int i = 3; i >= 0; i--) {
		for (int j = 0; j < 4; j++)
			printf("%c", g[j][i] ?
 '#' : '.');
		printf("\n");
	}
	printf("\n");
}
/************************/

const int maxn = 1005;

int N, M, P, B[maxn], G[15][15];
char order[maxn];

bool judge (int x, int y, const int t[4][2]) {
	for (int i = 0; i < 4; i++) {
		int p = x + t[i][0];
		int q = y + t[i][1];
		if (G[p][q])
			return false;
	}
	return true;
}

void tag (int x, int y, const int t[4][2]) {
	for (int i = 0; i < 4; i++) {
		int p = x + t[i][0];
		int q = y + t[i][1];
		G[p][q] = 1;
	}
}

void play(int t) {
	int x = 4, y = 9, c = 0;
	while (P < M) {
		if (order[P] == 'w') {
			if (judge(x, y, T[t][(c + 1) % C[t]]))
				c = (c + 1) % C[t];
		} else if (order[P] == 'a') {
			if (judge(x - 1, y, T[t][c]))
				x = x - 1;
		} else if (order[P] == 's') {
			if (judge(x, y - 1, T[t][c]))
				y = y - 1;
		} else if (order[P] == 'd') {
			if (judge(x + 1, y, T[t][c]))
				x = x + 1;
		}
		P++;

		if (!judge(x, y - 1, T[t][c]))
			break;
		y = y - 1;
	}
	tag(x, y, T[t][c]);
}

int remove () {
	int ret = 0;
	for (int j = 1; j <= 9; j++) {
		bool flag = true;
		for (int i = 1; i <= 9; i++) {
			if (G[i][j] == 0) {
				flag = false;
				break;
			}
		}

		if (flag) {
			ret++;
			for (int x = j; x < 12; x++) {
				for (int i = 1; i <= 9; i++)
					G[i][x] = G[i][x+1];
			}
			j--;
		}
	}
	return ret;
}

int solve () {
	scanf("%d%s", &N, order);
	P = 0;
	M = strlen(order);

	memset(G, 0, sizeof(G));
	for (int i = 0; i < 15; i++)
		G[i][0] = G[0][i] = G[10][i] = -1;

	int ret = 0, t;
	for (int i = 1; i <= N; i++) {
		scanf("%d", &t);
		play(t);
		ret += remove();
	}
	return ret;
}

int main () {
	int cas;
	scanf("%d", &cas);
	for (int kcas = 1; kcas <= cas; kcas++) {
		printf("Case %d: %d\n", kcas, solve ());
	}
	return 0;
}