poj 1038 - Bugs Integrated, Inc.(状态压缩dp)

const int tri[15] = {0, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049};
int n, m, k, dp[2][59049], pre[15], now[15];  //滚动数组来记录
bool g[155][15];
int getTen(int *a)//把一个用数组表示的状态转化为一个10进制整数
{
int ans = 0;
for(int i = 1; i <= m; ++i)
ans += a[i]*tri[i];
return ans;
}
void getTri(int v, int *a)//把一个整数转化为一个数组来表示状态
{
for(int i = 1; i <= m; ++i) {
a[i] = v%3;
v /= 3;
}
}
void dfs(int x, int y, int last, int key)
{
int k;
dp[x&1][key] = max(dp[x&1][key], last);
if(y >= m) return;
if(!pre[y] && !pre[y+1] && !now[y] && !now[y+1]) {//竖着切割
now[y] = now[y+1] = 2;
k = getTen(now);
dfs(x, y+2, last+1, k);
now[y] = now[y+1] = 0;
}
if(y<m-1 && !now[y] && !now[y+1] && !now[y+2]) {  //横着切割

now[y] = now[y+1] = now[y+2] = 2;
k = getTen(now);
dfs(x, y+3, last+1, k);
now[y] = now[y+1] = now[y+2] = 0;
}
dfs(x, y+1, last, key); //不做改动，直接下一行
}
int main()
{
int t, a, b, tmp;
scanf("%d", &t);
while(t--) {
scanf("%d%d%d", &n, &m, &k);
memset(g, 0, sizeof(g));
for(int i = 0; i < tri[m+1]; ++i) dp[1][i] = -1;
for(int i = 0; i < k; ++i) {
scanf("%d%d", &a, &b);
g[a][b] = 1;
}
for(int i = 1; i <= m; ++i)
pre[i] = g[1][i]+1; //设置第一行的状态，第0行的方块全部视为不可用
tmp = getTen(pre);
dp[1][tmp] = 0;
for(int i = 2; i <= n; ++i) {
for(int j = 0; j < tri[m+1]; ++j) dp[i&1][j] = -1;
for(int j = 0; j < tri[m+1]; ++j) {
if(dp[(i+1)&1][j] == -1) continue; //跳过不可能的子状态
getTri(j, pre);
for(int l = 1; l <= m; ++l)  //根据上一行的状态得到本行的状态
if(g[i][l]) now[l] = 2;
else now[l] = max(pre[l]-1, 0);
tmp = getTen(now);
dfs(i, 1, dp[(i+1)&1][j], tmp);
}
}
int ans = 0;
for(int i = 0; i < tri[m+1]; ++i)
ans = max(ans, dp[n&1][i]);
printf("%d\n", ans);
}
return 0;
}