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  • BZOJ 2763 飞行路线 BFS分层

    题目链接:

    https://www.lydsy.com/JudgeOnline/problem.php?id=2763

    题目大意:

    Alice和Bob现在要乘飞机旅行,他们选择了一家相对便宜的航空公司。该航空公司一共在n个城市设有业务,设这些城市分别标记为0到n-1,一共有m种航线,每种航线连接两个城市,并且航线有一定的价格。Alice和Bob现在要从一个城市沿着航线到达另一个城市,途中可以进行转机。航空公司对他们这次旅行也推出优惠,他们可以免费在最多k种航线上搭乘飞机。那么Alice和Bob这次出行最少花费多少?

    思路:

    BFS分层即可。

     1 #include<bits/stdc++.h>
     2 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
     3 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数,会超时
     4 #define Min(a, b) ((a) < (b) ? (a) : (b))
     5 #define Mem(a) memset(a, 0, sizeof(a))
     6 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
     7 #define MID(l, r) ((l) + ((r) - (l)) / 2)
     8 #define lson ((o)<<1)
     9 #define rson ((o)<<1|1)
    10 #define Accepted 0
    11 #pragma comment(linker, "/STACK:102400000,102400000")//栈外挂
    12 using namespace std;
    13 inline int read()
    14 {
    15     int x=0,f=1;char ch=getchar();
    16     while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    17     while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    18     return x*f;
    19 }
    20 typedef long long ll;
    21 const int maxn = 100000 + 10;
    22 const int MOD = 1000000007;//const引用更快,宏定义也更快
    23 const int INF = 1e9 + 7;
    24 const double eps = 1e-6;
    25 bool vis[maxn][11];
    26 struct edge
    27 {
    28     int v, w;
    29     edge(){}
    30     edge(int v, int w):v(v), w(w){}
    31 };
    32 vector<edge>G[maxn];
    33 int n, m, k;
    34 int s, t;
    35 struct Heapnode
    36 {
    37     int d, id, k;//距离 点 层数
    38     Heapnode(){}
    39     Heapnode(int d, int id, int k):d(d), id(id), k(k){}
    40     bool operator< (const Heapnode & a)const
    41     {
    42         return d > a.d || d == a.d && k > a.k;
    43     }
    44 };
    45 priority_queue<Heapnode>q;
    46 int BFS()
    47 {
    48     q.push(Heapnode(0, s, 0));
    49     while(!q.empty())
    50     {
    51         Heapnode now = q.top();
    52         q.pop();
    53         if(vis[now.id][now.k])continue;
    54         vis[now.id][now.k] = 1;
    55         if(now.id == t)
    56         {
    57             return now.d;
    58         }
    59         for(int i = 0; i < G[now.id].size(); i++)
    60         {
    61             int v = G[now.id][i].v;
    62             int w = G[now.id][i].w;
    63             if(!vis[v][now.k])
    64             {
    65                 q.push(Heapnode(now.d + w, v, now.k));
    66             }
    67             if(!vis[v][now.k + 1] && now.k + 1 <= k)
    68             {
    69                 q.push(Heapnode(now.d, v, now.k + 1));
    70             }
    71         }
    72     }
    73 }
    74 int main()
    75 {
    76     scanf("%d%d%d", &n, &m, &k);
    77     scanf("%d%d", &s, &t);
    78     while(m--)
    79     {
    80         int u, v, w;
    81         scanf("%d%d%d", &u, &v, &w);
    82         G[u].push_back(edge(v, w));
    83         G[v].push_back(edge(u, w));
    84     }
    85     printf("%d
    ", BFS());
    86     return Accepted;
    87 }
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  • 原文地址:https://www.cnblogs.com/fzl194/p/9711269.html
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