pid=5374">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;

}