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\n", BFS());
86     return Accepted;
87 }

 

posted @ 2018-09-27 09:07  _努力努力再努力x  阅读(174)  评论(0编辑  收藏  举报