USACO17FEB Why Did the Cow Cross the Road I G

题目传送门

难得不卡(SPFA)

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因为可爱的贝茜会走三步停一步，所以可以将这三步并为一步来考虑，花费的时间为t * 3 + cost[x][y]
需要注意的是，贝茜可以走一步再走回来，所以她周围的点也要考虑到，这样一共是16个点
另外，当贝茜到达终点时不一定正好走三步，所以要将和终点距离为(1,2)的点也连向终点

建图结束后，跑一遍最短路就可以了

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#define LL long long
using namespace std;
LL read() {
    LL k = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9')
      k = k * 10 + c - 48, c = getchar();
    return k * f;
}
struct zzz {
    int t, nex, len;
}e[500010 << 1]; int head[100010], tot;
void add(int x, int y, int z) {
    e[++tot].t = y;
    e[tot].len = z;
    e[tot].nex = head[x];
    head[x] = tot;
}
int mapp[110][110], n, t;
inline int calc(int x, int y) {
    return (x-1) * n + y;
}
inline bool judge(int x, int y) {
    return x >= 1 && x <= n && y >= 1 && y <= n;
}
int fx[20] = {0, -2, -1, 1, 2, 2, 1, -1, -2, 0, 1, 0, -1, 0, 3, 0, -3};
int fy[20] = {0, 1, 2, 2, 1, -1, -2, -2, -1, 1, 0, -1, 0, 3, 0, -3, 0};
bool vis[100010]; int dis[100010];
void SPFA(int s) {
    memset(dis, 127, sizeof(dis)); dis[s] = 0; vis[s] = 1;
    queue <int> q; q.push(s);
    while(!q.empty()) {
        int k = q.front(); q.pop(); vis[k] = 0;
        for(int i = head[k]; i; i = e[i].nex) {
            if(dis[e[i].t] > dis[k] + e[i].len) {
                dis[e[i].t] = dis[k] + e[i].len;
                if(!vis[e[i].t])
                    vis[e[i].t] = 1, q.push(e[i].t);
            }
        }
    }
}
int main() {
    n = read(), t = read();
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j) mapp[i][j] = read();
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j) {
            for(int k = 1; k <= 16; ++k) {
                int xx = i + fx[k], yy = j + fy[k];
                if(judge(xx, yy))
                    add(calc(i, j), calc(xx, yy), 3 * t + mapp[xx][yy]);
            }
        }
    add(calc(n-1, n-1), calc(n, n), t << 1);
    add(calc(n-2, n), calc(n, n), t << 1);
    add(calc(n, n-2), calc(n, n), t << 1);
    add(calc(n-1, n), calc(n, n), t);
    add(calc(n, n-1), calc(n, n), t);

    SPFA(calc(1, 1));

    printf("%d
", dis[calc(n, n)]);
    return 0;
}