[BZOJ4992] [Usaco2017 Feb]Why Did the Cow Cross the Road（spfa）

传送门

把每个点和曼哈顿距离距离它3步或1步的点连一条边，边权为3 * t + a[x][y]

因为，走3步，有可能是3步，也有可能是1步（其中一步拐了回来）

最后，把终点和曼哈顿距离距离它1步和2布的点连一条边，边权为 步数 * t

跑一边spfa即可

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#define N 1000001
#define idx(i, j) ((i - 1) * n + j)

using namespace std;

int n, t, cnt;
int a[101][101], d[4][N][2], tot[4];
int head[N], to[N], next[N], val[N], dis[N];
bool vis[N];
queue <int> q;

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void add(int x, int y, int z)
{
	to[cnt] = y;
	val[cnt] = z;
	next[cnt] = head[x];
	head[x] = cnt++;
}

inline void spfa()
{
	int i, u, v;
	memset(dis, 127, sizeof(dis));
	dis[1] = 0;
	q.push(1);
	while(!q.empty())
	{
		u = q.front();
		q.pop();
		vis[u] = 0;
		for(i = head[u]; ~i; i = next[i])
		{
			v = to[i];
			if(dis[v] > dis[u] + val[i])
			{
				dis[v] = dis[u] + val[i];
				if(!vis[v])
				{
					q.push(v);
					vis[v] = 1;
				}
			}
		}
	}
}

int main()
{
	int i, j, k, l, x, y;
	n = read();
	t = read();
	for(i = 1; i <= 3; i++)
		for(j = 0; j <= i; j++)
		{
			d[i][++tot[i]][0] = j, d[i][tot[i]][1] = i - j;
			d[i][++tot[i]][0] = j, d[i][tot[i]][1] = -i + j;
			d[i][++tot[i]][0] = -j, d[i][tot[i]][1] = i - j;
			d[i][++tot[i]][0] = -j, d[i][tot[i]][1] = -i + j;
		}
	memset(head, -1, sizeof(head));
	for(i = 1; i <= n; i++)
		for(j = 1; j <= n; j++)
			a[i][j] = read();
	for(i = 1; i <= n; i++)
		for(j = 1; j <= n; j++)
			for(l = 1; l <= 3; l += 2)
				for(k = 1; k <= tot[l]; k++)
				{
					x = i + d[l][k][0];
					y = j + d[l][k][1];
					if(1 <= x && x <= n && 1 <= y && y <= n)
						add(idx(i, j), idx(x, y), 3 * t + a[x][y]);
				}
	for(i = 1; i <= 2; i++)
		for(j = 1; j <= tot[i]; j++)
		{
			x = n + d[i][j][0];
			y = n + d[i][j][1];
			if(1 <= x && x <= n && 1 <= y && y <= n)
				add(idx(x, y), n * n, i * t);
		}
	spfa();
	printf("%d\n", dis[n * n]);
	return 0;
}