Made In Heaven
题目描述
One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are N spots in the jail and M roads connecting some of the spots. JOJO finds that Pucci knows the route of the former (K-1)-th shortest path. If Pucci spots JOJO in one of these K-1 routes, Pucci will use his stand Whitesnake and put the disk into JOJO’s body, which means JOJO won’t be able to make it to the destination. So, JOJO needs to take the K-th quickest path to get to the destination. What’s more, JOJO only has T units of time, so she needs to hurry.
JOJO starts from spot S, and the destination is numbered E. It is possible that JOJO’s path contains any spot more than one time. Please tell JOJO whether she can make arrive at the destination using no more than T units of time.
输入
There are at most 50 test cases.
The first line contains two integers N and M (1≤N≤1000,0≤M≤10000). Stations are numbered from 1 to N.
The second line contains four numbers S, E, Kand T( 1≤S,E≤N,S≠E,1≤K≤10000, 1≤T≤100000000).
Then M lines follows, each line containing three numbers U, V and W (1≤U,V≤N,1≤W≤1000) . It shows that there is a directed road from U-th spot to V-th spot with time W.
It is guaranteed that for any two spots there will be only one directed road from spot A to spot B(1≤A,B≤N,A≠B), but it is possible that both directed road <A,B> and directed road <B,A> exist.
All the test cases are generated randomly.
输出
One line containing a sentence. If it is possible for JOJO to arrive at the destination in time, output “yareyaredawa” (without quote), else output “Whitesnake!” (without quote).
样例输入
2 2
1 2 2 14
1 2 5
2 1 4
样例输出
yareyaredawa
题意+题解
第K短路 A* 模板题
关于A*算法及模板来自https://blog.csdn.net/z_mendez/article/details/47057461
代码
#include<bits/stdc++.h>
using namespace std;
#define INF 0xffffff
#define MAXN 100010
struct node
{
int to;
int val;
int next;
};
struct node2
{
int to;
int g,f;
bool operator<(const node2 &r ) const
{
if(r.f==f)
return r.g<g;
return r.f<f;
}
};
node edge[MAXN],edge2[MAXN];
int n,m,s,t,k,T,cnt,cnt2,ans;
int dis[1010],visit[1010],head[1010],head2[1010];
void init()
{
memset(head2,-1,sizeof(head2));
memset(head,-1,sizeof(head));
cnt=cnt2=1;
}
void addedge(int from,int to,int val)
{
edge[cnt].to=to;
edge[cnt].val=val;
edge[cnt].next=head[from];
head[from]=cnt++;
}
void addedge2(int from,int to,int val)
{
edge2[cnt2].to=to;
edge2[cnt2].val=val;
edge2[cnt2].next=head2[from];
head2[from]=cnt2++;
}
bool spfa(int s,int n,int head[],node edge[],int dist[])
{
queue<int>Q1;
int inq[1010];
for(int i=0;i<=n;i++)
{
dis[i]=INF;
inq[i]=0;
}
dis[s]=0;
Q1.push(s);
inq[s]++;
while(!Q1.empty())
{
int q=Q1.front();
Q1.pop();
inq[q]--;
if(inq[q]>n)
return false;
int k=head[q];
while(k>=0)
{
if(dist[edge[k].to]>dist[q]+edge[k].val)
{
dist[edge[k].to]=edge[k].val+dist[q];
if(!inq[edge[k].to])
{
inq[edge[k].to]++;
Q1.push(edge[k].to);
}
}
k=edge[k].next;
}
}
return true;
}
int A_star(int s,int t,int n,int k,int head[],node edge[],int dist[])
{
node2 e,ne;
int cnt=0;
priority_queue<node2>Q;
if(s==t)
k++;
if(dis[s]==INF)
return -1;
e.to=s;
e.g=0;
e.f=e.g+dis[e.to];
Q.push(e);
while(!Q.empty())
{
e=Q.top();
Q.pop();
if(e.to==t)//找到一条最短路径
{
cnt++;
}
if(cnt==k)//找到k短路
{
return e.g;
}
for(int i=head[e.to]; i!=-1; i=edge[i].next)
{
ne.to=edge[i].to;
ne.g=e.g+edge[i].val;
ne.f=ne.g+dis[ne.to];
Q.push(ne);
}
}
return -1; // 找不到
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
scanf("%d%d%d%d",&s,&t,&k,&T);
for(int i=1;i<=m;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);
addedge2(b,a,c);
}
spfa(t,n,head2,edge2,dis);
ans=A_star(s,t,n,k,head,edge,dis);
//printf("%d
",ans);
if(ans==-1||ans > T) printf("Whitesnake!
");
else printf("yareyaredawa
");
}
return 0;
}