USACO 2017 FEB Gold visitfj 最短路

　　题意

　　　　有一幅n*n的方格图，n <= 100，每个点上有一个值。从（1,1）出发，走到（n,n），只能走四联通。每走一步花费t，每走三步需要花费走完三步后到达格子的值。求最小花费的值。

　　拆点，dis[i][j]表示到达第i个点时走的总步数模3等于j时的最小花费值。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>

using namespace std;

const int maxn = ;
const int INF = 0x3fffffff;
int n, t, w[maxn][maxn];
struct Node
{
    int x, y, z;
    Node (int x = , int y = , int z = ):
        x(x), y(y), z(z) {}
};
queue <Node> q;
int dx[] = {, , , -}, dy[] = {, -, , };
int dis[maxn][maxn][];
bool vis[maxn][maxn][];

int main()
{
    freopen("visitfj.in", "r", stdin);
    freopen("visitfj.out", "w", stdout);
    scanf("%d %d", &n, &t);
    for (int i = ; i <= n; ++i)
        for (int j = ; j <= n; ++j)
            scanf("%d", &w[i][j]);
    for (int i = ; i <= n; ++i)
        for (int j = ; j <= n; ++j)
            for (int k = ; k < ; ++k)
                dis[i][j][k] = INF, vis[i][j][k] = false;
    dis[][][] = ;
    vis[][][] = true;
    q.push(Node(, , ));
    while (!q.empty())
    {
        Node now = q.front();
        vis[now.x][now.y][now.z] = false;
        q.pop();
        for (int i = ; i < ; ++i)
        {
            int x = now.x+dx[i], y = now.y+dy[i], z = (now.z+)%;
            if (x <  || y <  || x > n || y > n)
                continue ;
            if (dis[x][y][z] > dis[now.x][now.y][now.z]+t+((z == ) ? w[x][y] : ))
            {
                dis[x][y][z] = dis[now.x][now.y][now.z]+t+((z == ) ? w[x][y] : );
                if (!vis[x][y][z])
                    vis[x][y][z] = true, q.push(Node(x, y, z));
            }
        }
    }
    printf("%d\n", min(dis[n][n][], min(dis[n][n][], dis[n][n][])));
    return ;
}