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?
多加一维进去dijkstra就可以了。。好像usaco有过。。
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-‘0‘,c=getchar();
return x;
}
const int nmax=1e4+5;
const int maxn=1e5+5;
const int inf=0x7f7f7f7f;
struct edge{
int to,dist;edge *next;
};
edge es[maxn],*pt=es,*head[nmax];
void add(int u,int v,int d){
pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}
int dist[nmax][11];
struct node{
int x,k,dist;
node(int x,int k,int dist):x(x),k(k),dist(dist){};
node(){};
bool operator<(const node&rhs)const{
return dist>rhs.dist;}
};
priority_queue<node>q;
void dijkstra(int s,int t,int k){
clr(dist,0x7f);dist[s][0]=0;q.push(node(s,0,0));
node o;int ta,td,tk,to;
while(!q.empty()){
o=q.top();q.pop();ta=o.x,td=o.dist,tk=o.k;
if(dist[ta][tk]!=td) continue;
qwq(ta){
to=o->to;
if(dist[o->to][tk]>td+o->dist){
dist[o->to][tk]=td+o->dist;q.push(node(o->to,tk,td+o->dist));
}
if(tk<k&&dist[o->to][tk+1]>td){
dist[o->to][tk+1]=td;q.push(node(o->to,tk+1,td));
}
}
}
int ans=inf;
rep(i,0,k) ans=min(ans,dist[t][i]);
printf("%d\n",ans);
return ;
}
int main(){
int n=read(),m=read(),k=read(),u,v,d,s=read(),t=read();
rep(i,1,m) u=read(),v=read(),d=read(),add(u,v,d);
dijkstra(s,t,k);
return 0;
}
对于30%的数据,2<=n<=50,1<=m<=300,k=0;
对于50%的数据,2<=n<=600,1<=m<=6000,0<=k<=1;
对于100%的数据,2<=n<=10000,1<=m<=50000,0<=k<=10.
原文:http://www.cnblogs.com/fighting-to-the-end/p/5885032.html