BZOJ 2763 飞行路线 BFS分层

时间:2021-11-26 11:09:07

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。

 #include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
#define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数,会超时
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Mem(a) memset(a, 0, sizeof(a))
#define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
#define MID(l, r) ((l) + ((r) - (l)) / 2)
#define lson ((o)<<1)
#define rson ((o)<<1|1)
#define Accepted 0
#pragma comment(linker, "/STACK:102400000,102400000")//栈外挂
using namespace std;
inline int read()
{
int x=,f=;char ch=getchar();
while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
typedef long long ll;
const int maxn = + ;
const int MOD = ;//const引用更快,宏定义也更快
const int INF = 1e9 + ;
const double eps = 1e-;
bool vis[maxn][];
struct edge
{
int v, w;
edge(){}
edge(int v, int w):v(v), w(w){}
};
vector<edge>G[maxn];
int n, m, k;
int s, t;
struct Heapnode
{
int d, id, k;//距离 点 层数
Heapnode(){}
Heapnode(int d, int id, int k):d(d), id(id), k(k){}
bool operator< (const Heapnode & a)const
{
return d > a.d || d == a.d && k > a.k;
}
};
priority_queue<Heapnode>q;
int BFS()
{
q.push(Heapnode(, s, ));
while(!q.empty())
{
Heapnode now = q.top();
q.pop();
if(vis[now.id][now.k])continue;
vis[now.id][now.k] = ;
if(now.id == t)
{
return now.d;
}
for(int i = ; i < G[now.id].size(); i++)
{
int v = G[now.id][i].v;
int w = G[now.id][i].w;
if(!vis[v][now.k])
{
q.push(Heapnode(now.d + w, v, now.k));
}
if(!vis[v][now.k + ] && now.k + <= k)
{
q.push(Heapnode(now.d, v, now.k + ));
}
}
}
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
scanf("%d%d", &s, &t);
while(m--)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
G[u].push_back(edge(v, w));
G[v].push_back(edge(u, w));
}
printf("%d\n", BFS());
return Accepted;
}