ECNU 2018 10月月赛 E 盖房子 (bitset + 倍增)

题目链接  ECNU Monthly 2018.10 Problem E

从开场写到结束……

显然要把三角形分成上下两部分。

把每一部分分成三部分,以上部分为例。

上面和右边,以及左下角的正方形。

也就是两个小三角形和一个正方形合起来。

处理正方形的时候稍微麻烦一些。

然后直接倍增就可以了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
#define fi first
#define se second
#define MP make_pair typedef long long LL; const int N = 1e3 + 10; bitset <102> f[N][N][10], g[N][N][10], ff[N][N][10], gg[N][N][10];
int n, m, q;
int T;
int lg[N + 10]; inline int check(int x, int y){
return x >= 1 && x <= n && y >= 1 && y <= m;
} int main(){ lg[1] = 0;
rep(i, 2, 1001) lg[i] = lg[i >> 1] + 1;
scanf("%d%d", &n, &m); rep(i, 1, n){
rep(j, 1, m){
int x;
scanf("%d", &x);
f[i][j][0].set(x);
g[i][j][0].set(x);
ff[i][j][0].set(x);
gg[i][j][0].set(x);
}
} rep(k, 1, 9){
rep(i, 1, n){
rep(j, 1, m){
int xx, yy, zz = 1 << (k - 1);
f[i][j][k] |= f[i][j][k - 1];
xx = i - zz;
yy = j; if (check(xx, yy)) f[i][j][k] |= f[xx][yy][k - 1]; xx = i;
yy = j + zz;
if (check(xx, yy)) f[i][j][k] |= f[xx][yy][k - 1]; xx = i - zz;
yy = j + zz;
if (check(xx, yy)) f[i][j][k] |= f[xx][yy][k - 1]; ff[i][j][k] |= ff[i][j][k - 1]; xx = i + zz;
yy = j;
if (check(xx, yy)) ff[i][j][k] |= ff[xx][yy][k - 1]; xx = i;
yy = j + zz;
if (check(xx, yy)) ff[i][j][k] |= ff[xx][yy][k - 1]; xx = i + zz;
yy = j + zz;
if (check(xx, yy)) ff[i][j][k] |= ff[xx][yy][k - 1]; }
}
} rep(k, 1, 9){
rep(i, 1, n){
rep(j, 1, m){
int xx, yy, zz = 1 << (k - 1);
g[i][j][k] |= f[i][j][k - 1];
xx = i - zz;
yy = j;
if (check(xx, yy)) g[i][j][k] |= g[xx][yy][k - 1]; xx = i;
yy = j + zz;
if (check(xx, yy)) g[i][j][k] |= g[xx][yy][k - 1]; gg[i][j][k] |= ff[i][j][k - 1];
xx = i + zz;
yy = j;
if (check(xx, yy)) gg[i][j][k] |= gg[xx][yy][k - 1]; xx = i;
yy = j + zz;
if (check(xx, yy)) gg[i][j][k] |= gg[xx][yy][k - 1]; }
}
} scanf("%d", &q);
while (q--){
int x, y, z;
scanf("%d%d%d", &x, &y, &z); if (z == 1){
puts("1");
continue;
} int c = lg[z]; bitset <102> ret; int xx, yy, zz = 1 << c; xx = x - z + zz;
yy = y; ret |= g[xx][yy][c]; xx = x;
yy = y + z - zz; ret |= g[xx][yy][c]; int t = (z) >> 1; int l = lg[t], ll = 1 << l; ret |= f[x][y][l]; xx = x - t + ll;
yy = y;
ret |= f[xx][yy][l]; xx = x;
yy = y + t - ll;
ret |= f[xx][yy][l]; xx = x - t + ll;
yy = y + t - ll;
ret |= f[xx][yy][l]; xx = x + z - zz;
yy = y;
ret |= gg[xx][yy][c]; xx = x;
yy = y + z - zz;
ret |= gg[xx][yy][c]; ret |= ff[x][y][l];
xx = x + t - ll;
yy = y;
ret |= ff[xx][yy][l]; xx = x;
yy = y + t - ll;
ret |= ff[xx][yy][l]; xx = x + t - ll;
yy = y + t - ll;
ret |= ff[xx][yy][l]; printf("%d\n", (int)ret.count());
} return 0; }
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