Blog of RuSun

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

P4009 汽车加油行驶问题

P4009 汽车加油行驶问题

分层图最短路,模拟过程即可。

查看代码
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#include <cstdio>
#include <queue>
using namespace std;
typedef long long LL;
void chkmin(LL &x, LL k)
{
(x > k) && (x = k);
}
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');
}
LL inf = 1e18;
const int N = 110, K = 20;
const int dx[] = {0, -1, 0, 1}, dy[] = {-1, 0, 1, 0};
bool mp[N][N], vis[N][N][K];
int n, m, A, B, C;
LL d[N][N][K];
bool inside(int x, int y)
{
return x > 0 && y > 0 && x <= n && y <= n;
}
struct Node
{
int x, y, t;
LL s;
friend bool operator<(const Node &x, const Node &y)
{
return x.s > y.s;
}
};
int main()
{
read(n), read(m), read(A), read(B), read(C);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
read(mp[i][j]);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 0; k <= m; k++)
d[i][j][k] = inf;
d[1][1][m] = 0;
priority_queue<Node> q;
q.push((Node){1, 1, m, 0});
while (!q.empty())
{
Node t = q.top();
q.pop();
if (vis[t.x][t.y][t.t] || !t.t)
continue;
vis[t.x][t.y][t.t] = true;
for (int i = 0; i < 4; i++)
{
int nx = t.x + dx[i], ny = t.y + dy[i];
if (!inside(nx, ny))
continue;
if (mp[nx][ny])
{
if (t.s + A + (i < 2) * B < d[nx][ny][m])
{
d[nx][ny][m] = t.s + A + (i < 2) * B;
q.push((Node){nx, ny, m, d[nx][ny][m]});
}
continue;
}
if (t.s + (i < 2) * B < d[nx][ny][t.t - 1])
{
d[nx][ny][t.t - 1] = t.s + (i < 2) * B;
q.push((Node){nx, ny, t.t - 1, d[nx][ny][t.t - 1]});
}
if (!(nx == n && ny == n) && t.s + A + C + (i < 2) * B < d[nx][ny][m])
{
d[nx][ny][m] = t.s + A + C + (i < 2) * B;
q.push((Node){nx, ny, m, d[nx][ny][m]});
}
}
}
LL res = inf;
for (int i = 0; i <= m; i++)
chkmin(res, d[n][n][i]);
write(res);
return 0;
}