【USACO 2017Feb】 Why Did the Cow Cross the Road

【题目链接】

           点击打开链接

【算法】

         dist[i][j][k]表示当前走到(i,j)，走的步数除以3的余数为k的最小花费

         spfa即可

【代码】

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 110
const int INF = 1e9;

struct info
{
        int x,y,s;
};

const int dx[4] = {0,0,-1,1};
const int dy[4] = {-1,1,0,0};

int i,j,n,t;
int val[MAXN][MAXN];

template <typename T> inline void read(T &x)
{
    int f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x/10);
    putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}
bool ok(int x,int y)
{
        return x >= 1 && x <= n && y >= 1 && y <= n;
}
inline void spfa()
{
        int i,j,tx,ty,ans;
        queue< info > q;
        static int dist[MAXN][MAXN][3],inq[MAXN][MAXN][3];
        info cur;
        for (i = 1; i <= n; i++)
        {
                for (j = 1; j <= n; j++)
                {
                        dist[i][j][0] = dist[i][j][1] = dist[i][j][2] = INF;
                }
        }
        dist[1][1][0] = 0;
        inq[1][1][0] = 1;
        q.push((info){1,1,0});
        while (!q.empty())
        {
                cur = q.front(); 
                inq[cur.x][cur.y][cur.s] = 0;
                q.pop();    
                for (i = 0; i < 4; i++)
                {
                        tx = cur.x + dx[i];
                        ty = cur.y + dy[i];
                        if (ok(tx,ty))
                        {
                                if (!cur.s)
                                {
                                        if (dist[cur.x][cur.y][0] + t < dist[tx][ty][1])
                                        {
                                                dist[tx][ty][1] = dist[cur.x][cur.y][0] + t;
                                                if (!inq[tx][ty][1])
                                                {
                                                        inq[tx][ty][1] = 1;
                                                        q.push((info){tx,ty,1});
                                                }
                                        }
                                }
                                if (cur.s == 1)
                                {
                                        if (dist[cur.x][cur.y][1] + t < dist[tx][ty][2])
                                        {
                                                dist[tx][ty][2] = dist[cur.x][cur.y][1] + t;
                                                if (!inq[tx][ty][2])
                                                {
                                                        inq[tx][ty][2] = 1;
                                                        q.push((info){tx,ty,2});
                                                }
                                        }
                                }
                                if (cur.s == 2)
                                {
                                        if (dist[cur.x][cur.y][2] + val[tx][ty] + t < dist[tx][ty][0])
                                        {
                                                dist[tx][ty][0] = dist[cur.x][cur.y][2] + val[tx][ty] + t;
                                                if (!inq[tx][ty][0])
                                                {
                                                        inq[tx][ty][0] = 1;
                                                        q.push((info){tx,ty,0});
                                                }
                                        }
                                }
                        }
                }
        }        
        ans = min(min(dist[n][n][0],dist[n][n][1]),dist[n][n][2]);
        writeln(ans);
}

int main() {
        
        read(n); read(t);
        
        for (i = 1; i <= n; i++)
        {
                for (j = 1; j <= n; j++)
                {
                        read(val[i][j]);        
                }        
        }
        spfa();
        
        return 0;
    
}