[CERC2016]机棚障碍 Hangar Hurdles
传送门:2016-2017 ACM-ICPC, Central Europe Regional Contest (CERC 16)
2016-2017年中欧地区的比赛题，不得不说算是冷门了...
题面大意：
给定\(n*m\)网格图，某些位置为·障碍，多组询问，询问从给定位置到达给定位置间可以推过的最大箱子。
题目里有些需要注意的，“从某点”意味着这个点作为箱子的中心点，那么显然箱子的大小只能为奇数。
分析：
可以说是典型的缝合怪了......，是口头\(AC\)五分钟实际处理两小时的类型。我们考虑将每个位置可通过的最大箱子，然后建图，构造最大生成树，询问时就先判断是否在同一生成树中，若在同一生成树中就用\(LCA\)维护快速求出两点间路径的最小值。
首先考虑如何处理出每个位置可通过的最大箱子，前缀和+二分即可。
bool check(int x, int y, int z)
{
if (x + z > n || y + z > n || x - z < 1 || y - z < 1)
return false;
if (f[x + z][y + z] - f[x + z][y - z - 1] - f[x - z - 1][y + z] + f[x - z - 1][y - z - 1] == (2 * z + 1) * (2 * z + 1))
return true;
return false;
}
int work(int x, int y)
{
int l = 0, r = n, ans;
while (l <= r)
{
int mid = (l + r) >> 1;
if (check(x, y, mid))
{
ans = mid;
l = mid + 1;
}
else
{
r = mid - 1;
}
}
return ans;
}
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (a[i][j] == '.')
{
qaq++;
}
f[i][j] = f[i - 1][j] + f[i][j - 1] - f[i - 1][j - 1] + (a[i][j] == '.');
}
}
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (a[i][j] == '.')
{
qwq[i][j] = 2 * work(i, j) + 1;
}
}
}
然后就是建图和最大生成树，唯一可能需要注意的就是\(LCA\)的维护，具体可以看\(NOIP\)某年的货车运输，两者几乎相同。
void dfs(int x, int father, int col, int z)
{
vis[x] = col;
dep[x] = dep[father] + 1;
F[x][0] = father;
dis[x][0] = z;
for (int i = 1; i <= 20; ++i)
{
F[x][i] = F[F[x][i - 1]][i - 1];
dis[x][i] = min(dis[x][i - 1], dis[F[x][i - 1]][i - 1]);
}
for (auto u : G[x])
{
if (u.first != father)
{
dfs(u.first, x, col, u.second);
}
}
}
int lca(int x, int y)
{
int ans = 1005*1005;
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; --i)
{
if (dep[F[x][i]] >= dep[y])
{
ans = min(ans, dis[x][i]);
x = F[x][i];
}
}
if (x == y)
{
return ans;
}
for (int i = 20; i >= 0; --i)
{
if (F[x][i] != F[y][i])
{
ans = min(ans, min(dis[x][i], dis[y][i]));
x = F[x][i], y = F[y][i];
}
}
ans = min(ans, min(dis[x][0], dis[y][0]));
return ans;
}
理论上来讲缩点和树剖会让代码处理速度更快，但这样代码长度可能达到\(300\)行往上...所以就写了常数更大更容易写的倍增求\(LCA\)
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <string.h>
#include <cmath>
#include <deque>
#include <vector>
#define Lint long long
using namespace std;
typedef pair<int, int> P;
int n, q, m, qaq;
char a[1005][1005];
int f[1005][1005];
int qwq[1005][1005];
int F[1005*1005][21], dis[1005*1005][21];
int vis[1005 * 1005];
int dep[1005 * 1005];
vector<P> G[1005 * 1005];
struct node
{
int x, y, z;
} e[1005 * 1005 * 2];
int fa[1005 * 1005], sumcol;
bool cmp(node x, node y) { return x.z > y.z; }
bool check(int x, int y, int z)
{
if (x + z > n || y + z > n || x - z < 1 || y - z < 1)
return false;
if (f[x + z][y + z] - f[x + z][y - z - 1] - f[x - z - 1][y + z] + f[x - z - 1][y - z - 1] == (2 * z + 1) * (2 * z + 1))
return true;
return false;
}
int getid(int x, int y) { return (x - 1) * n + y; }
int Find(int x)
{
if (x == fa[x])
return x;
return fa[x] = Find(fa[x]);
}
int work(int x, int y)
{
int l = 0, r = n, ans;
while (l <= r)
{
int mid = (l + r) >> 1;
if (check(x, y, mid))
{
ans = mid;
l = mid + 1;
}
else
{
r = mid - 1;
}
}
return ans;
}
void krsual()
{
for (int i = 1; i <= m; ++i)
{
int le = Find(e[i].x), re = Find(e[i].y);
if (fa[le] != re)
{
fa[le] = re;
G[e[i].x].push_back({e[i].y, e[i].z});
G[e[i].y].push_back({e[i].x, e[i].z});
}
}
}
void dfs(int x, int father, int col, int z)
{
vis[x] = col;
dep[x] = dep[father] + 1;
F[x][0] = father;
dis[x][0] = z;
for (int i = 1; i <= 20; ++i)
{
F[x][i] = F[F[x][i - 1]][i - 1];
dis[x][i] = min(dis[x][i - 1], dis[F[x][i - 1]][i - 1]);
}
for (auto u : G[x])
{
if (u.first != father)
{
dfs(u.first, x, col, u.second);
}
}
}
int lca(int x, int y)
{
int ans = 1005*1005;
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; --i)
{
if (dep[F[x][i]] >= dep[y])
{
ans = min(ans, dis[x][i]);
x = F[x][i];
}
}
if (x == y)
{
return ans;
}
for (int i = 20; i >= 0; --i)
{
if (F[x][i] != F[y][i])
{
ans = min(ans, min(dis[x][i], dis[y][i]));
x = F[x][i], y = F[y][i];
}
}
ans = min(ans, min(dis[x][0], dis[y][0]));
return ans;
}
int main()
{
memset(dis, 0x3f, sizeof(dis));
scanf("%d", &n);
for (int i = 1; i <= n; ++i)
{
scanf("%s", a[i] + 1);
}
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (a[i][j] == '.')
{
qaq++;
}
f[i][j] = f[i - 1][j] + f[i][j - 1] - f[i - 1][j - 1] + (a[i][j] == '.');
}
}
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (a[i][j] == '.')
{
qwq[i][j] = 2 * work(i, j) + 1;
}
}
}
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (a[i][j] == '.')
{
if (i <= n - 1)
{
if (a[i + 1][j] == '.')
{
e[++m].x = getid(i, j);
e[m].y = getid(i + 1, j);
e[m].z = min(qwq[i][j], qwq[i + 1][j]);
}
}
if (j <= n - 1)
{
if (a[i][j + 1] == '.')
{
e[++m].x = getid(i, j);
e[m].y = getid(i, j + 1);
e[m].z = min(qwq[i][j], qwq[i][j + 1]);
}
}
}
}
}
sort(e + 1, e + 1 + m, cmp);
for (int i = 1; i <= n * n; ++i)
fa[i] = i;
krsual();
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (a[i][j] == '.')
{
if (!vis[getid(i, j)])
{
dfs(getid(i, j), 0, ++sumcol, 0);
}
}
}
}
// for(int i=0;i<=20;++i){
// cout<<F[getid(2,2)][i]<<" "<<dis[getid(2,2)][i]<<endl;
// }
scanf("%d", &q);
int bex, bey, rex, rey;
while (q--)
{
scanf("%d%d%d%d", &bex, &bey, &rex, &rey);
if (vis[getid(bex, bey)] != vis[getid(rex, rey)])
{
puts("0");
}
else
{
printf("%d\n", lca(getid(bex, bey), getid(rex, rey)));
}
}
return 0;
}
本文章为转载内容，我们尊重原作者对文章享有的著作权。如有内容错误或侵权问题，欢迎原作者联系我们进行内容更正或删除文章。
