HDU 5386 构造

#include <cstdio>
#include <cstring>
#define H (c[j] == 'H'?x[j]:i)
#define L (c[j] == 'L'?x[j]:i)
const int maxn = 5E2 + 10;
char c[maxn];
int T, n, m, inp[maxn][maxn], x[maxn], y[maxn], ans[maxn], cnt;
int main(int argc, char const *argv[])
{
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%d", &inp[i][j]);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%d", &inp[i][j]);
		getchar(); cnt = 0;
		for (int i = 1; i <= m; i++)
			scanf("%*c%c%d%d", &c[i], x + i, y + i);
		for (int k = m; k >= 1; k--)
			for (int j = 1; j <= m; j++)
				if (x[j])
				{
					bool ok = true;
					for (int i = 1; i <= n; i++)
						if (inp[H][L] && inp[H][L] != y[j])
						{ok = false; break;}
					if (ok)
					{
						for (int i = 1; i <= n; i++)
							inp[H][L] = 0;
						x[j] = 0; ans[cnt++] = j;
					}
				}
		for (int i = cnt - 1; i >= 0; i--) printf("%d%c", ans[i], i == 0 ? '\n' : ' ');
	}
	return 0;
}


题目保证一定有解,又因为操作是整行或整列替换且初始矩阵没有用(会覆盖),所以可以假设为零矩阵,然后从目标矩阵开始通过给的操作使整行或整列变为零,直到目标矩阵变为零矩阵,然后逆序输出操作序号,过程纯暴力

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