[luoguP2129] L国的战斗续之多路出击（模拟 || 矩阵）

传送门

1.模拟

easy

#include <cstdio>
#define N 500001

int n, m;
int X[N], Y[N], x[N], y[N], a = 1, b = 1, p, q;
char s[N][1];

int main()
{
    int i;
    scanf("%d %d", &n, &m);
    for(i = 1; i <= n; i++) scanf("%d %d", &X[i], &Y[i]);
    for(i = 1; i <= m; i++)
    {
        scanf("%s", s[i]);
        if(s[i][0] == 'm') scanf("%d %d", &x[i], &y[i]);
    }
    for(i = m; i >= 1; i--)
        if(s[i][0] == 'm')
        {
            p += x[i];
            q += y[i];
        }
        else if(s[i][0] == 'x')
        {
            a *= -1;
            p *= -1;
        }
        else
        {
            b *= -1;
            q *= -1;
        }
    for(i = 1; i <= n; i++) printf("%d %d
", X[i] * a + p, Y[i] * b + q);
    return 0;
}

　

2.矩阵优化

详见 Matrix67

#include <cstdio>
#include <cstring>
#define N 500001

int n, m;
int X[N], Y[N], x[N], y[N];
char s[N][2];

struct Matrix
{
	int n, m;
	int a[11][11];
	Matrix()
	{
		n = m = 0;
		memset(a, 0, sizeof(a));
	}
}sum;

inline Matrix operator * (const Matrix x, const Matrix y)
{
	Matrix ret;
	ret.n = x.n;
	ret.m = y.m;
	int i, j, k;
	for(i = 1; i <= x.n; i++)
		for(j = 1; j <= y.m; j++)
			for(k = 1; k <= y.n; k++)
				ret.a[i][j] += x.a[i][k] * y.a[k][j];
	return ret;
}

int main()
{
	int i;
	Matrix t;
	scanf("%d %d", &n, &m);
	sum.n = sum.m = 3;
	sum.a[1][1] = sum.a[2][2] = sum.a[3][3] = 1;
	for(i = 1; i <= n; i++) scanf("%d %d", &X[i], &Y[i]);
	for(i = 1; i <= m; i++)
	{
		scanf("%s", s[i]);
		if(s[i][0] == 'm') scanf("%d %d", &x[i], &y[i]);
	}
	for(i = m; i >= 1; i--)
	{
		memset(t.a, 0, sizeof(t.a));
		t.n = t.m = 3;
		t.a[1][1] = t.a[2][2] = t.a[3][3] = 1;
		if(s[i][0] == 'm') t.a[1][3] = x[i], t.a[2][3] = y[i];
		if(s[i][0] == 'x') t.a[1][1] = -1;
		if(s[i][0] == 'y') t.a[2][2] = -1;
		sum = t * sum;
	}
	for(i = 1; i <= n; i++)
	{
		memset(t.a, 0, sizeof(t.a));
		t.n = 3;
		t.m = t.a[3][1] = 1;
		t.a[1][1] = X[i];
		t.a[2][1] = Y[i];
		t = sum * t;
		printf("%d %d
", t.a[1][1], t.a[2][1]);
	}
	return 0;
}