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;
}