Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4011 孤岛营救问题

P4011 孤岛营救问题

不知道为什么网络流24题有这个题。

状压暴搜即可。

查看代码
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#include <cstdio>
#include <queue>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 20, M = 1 << 12;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
bool vis[N][N][M];
int n, m, p, g[N][N][4], key[N][N];
bool inside(int x, int y)
{
return x > 0 && y > 0 && x <= n && y <= m;
}
struct Node
{
int x, y, s, k;
};
queue<Node> q;
int main()
{
read(n), read(m), read(p), read(p);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
for (int k = 0; k < 4; k++)
{
int nx = i + dx[k], ny = j + dy[k];
inside(nx, ny) && (g[i][j][k] = -1);
}
for (int x1, y1, x2, y2, t; p; p--)
{
read(x1), read(y1), read(x2), read(y2), read(t);
for (int k = 0; k < 4; k++)
if (x1 + dx[k] == x2 && y1 + dy[k] == y2)
g[x1][y1][k] = t;
else if (x2 + dx[k] == x1 && y2 + dy[k] == y1)
g[x2][y2][k] = t;
}
read(p);
for (int x, y, t; p; p--)
{
read(x), read(y), read(t);
key[x][y] |= 1 << t;
}
q.push((Node){1, 1, 0, key[1][1]});
while (!q.empty())
{
Node t = q.front();
q.pop();
if (vis[t.x][t.y][t.k])
continue;
vis[t.x][t.y][t.k] = true;
if (t.x == n && t.y == m)
{
write(t.s);
return 0;
}
for (int k = 0; k < 4; k++)
{
int nx = t.x + dx[k], ny = t.y + dy[k];
int &u = g[t.x][t.y][k];
if (!inside(nx, ny) || !u || (u > 0 && !(t.k >> u & 1)))
continue;
q.push((Node){nx, ny, t.s + 1, t.k | key[nx][ny]});
}
}
write(-1);
return 0;
}