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; }

转载于:https://www.cnblogs.com/wzzkaifa/p/6805240.html