汽车加油行驶问题【网络流24题】【可以使用BFS】

题目链接

  这道题虽然说是网络流24题中的一题，但是我的第一想法确实去用BFS()跑一个最小的花费，但是由于加油的钱、向后走的钱、开设一个新的加油站的钱是不固定的，所以，我们需要进行相应的判断，跑所有可以达到终点的值去比较大小。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 105;
const int dir[4][2] =
{
    -1, 0,
    0, -1,
    0, 1,
    1, 0
};
int N, K, A, B, C, mp[maxN][maxN];
int vis[maxN][maxN][11];
struct node
{
    int x, y, step, cost;
    node(int a=0, int b=0, int c=0, int d=0):x(a), y(b), step(c), cost(d) {}
    friend bool operator < (node e1, node e2) { return e1.cost > e2.cost; }
};
queue<node> Q;
bool In_Map(int x, int y) { return x > 0 && y > 0 && x <= N && y <= N; }
inline int bfs()
{
    memset(vis, INF, sizeof(vis));
    while(!Q.empty()) Q.pop();
    Q.push(node(1, 1, K, 0));
    vis[1][1][K] = 0;
    int ans = INF;
    while(!Q.empty())
    {
        node tmp = Q.front(); Q.pop();
        for(int i=0, x, y, step, cost; i<4; i++)
        {
            x = tmp.x + dir[i][0];  y = tmp.y + dir[i][1];  step = tmp.step - 1;    cost = tmp.cost + (i < 2 ? B : 0);
            if(!In_Map(x, y)) continue;
            if(x == N && y == N) { ans = min(ans, cost); continue; }
            if(mp[x][y]) { step = K;   cost += A; }
            if(!step)
            {
                step = K; cost += A + C;
                if(vis[x][y][step] > cost)
                {
                    vis[x][y][step] = cost;
                    Q.push(node(x, y, step, cost));
                }
                continue;
            }
            if(vis[x][y][step] > cost)
            {
                vis[x][y][step] = cost;
                Q.push(node(x, y, step, cost));
            }
        }
    }
    return ans;
}
int main()
{
    scanf("%d%d%d%d%d", &N, &K, &A, &B, &C);
    for(int i=1; i<=N; i++) for(int j=1; j<=N; j++) scanf("%d", &mp[i][j]);
    printf("%d\n", bfs());
    return 0;
}