poj 2449 Remmarguts' Date (k短路模板)
Time Limit: 4000MS | Memory Limit: 65536K | |
Total Submissions: 30772 | Accepted: 8397 |
Description
"Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission."
"Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)"
Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister's help!
DETAILS: UDF's capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince' current place. M muddy directed sideways connect some of the stations. Remmarguts' path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate.
Input
The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).
Output
Sample Input
2 2
1 2 5
2 1 4
1 2 2
Sample Output
14 边可以重复走,
不严格的k短路
A*
估价函数为dis(起点,i)+dis(i,终点) 1、反向图上求出终点到每个点的最短路
2、起点入优先队列,
队首出队,
如果队首是终点,而且是第k次出队,
那么当前距离就是k短路
如果队首不是终点,便利与当前点连接的所有的点,入队 细节1:优先队列出入队不用vis数组判重,因为边可以重复走
细节2:如果第1步中,起点与终点不连通,输出-1结束,
否则进入A*,没有vis数组,出现环会死循环
细节3:如果起点=终点,令k++,因为起点会立即出队
#include<queue>
#include<cstdio>
#include<cstring>
#define N 1001
#define M 100001
using namespace std;
int n,s,t,k;
int dis1[N];
bool vis[N];
int front[N],to[M],nxt[M],val[M],tot;
int front2[N],to2[M],nxt2[M],val2[M],tot2;
struct node
{
int num,dis;
bool operator < (node p) const
{
return dis+dis1[num]>p.dis+dis1[p.num];
}
}now,nt;
void add(int u,int v,int w)
{
to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; val[tot]=w;
to2[++tot2]=u; nxt2[tot2]=front2[v]; front2[v]=tot2; val2[tot2]=w;
}
void init()
{
int m,u,v,w;
scanf("%d%d",&n,&m);
while(m--)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
}
scanf("%d%d%d",&s,&t,&k);
}
void spfa()
{
memset(dis1,,sizeof(dis1));
queue<int>q;
dis1[t]=;
vis[t]=true;
q.push(t);
int now;
while(!q.empty())
{
now=q.front();
q.pop();
vis[now]=false;
for(int i=front2[now];i;i=nxt2[i])
if(dis1[to2[i]]>dis1[now]+val2[i])
{
dis1[to2[i]]=dis1[now]+val2[i];
if(!vis[to2[i]])
{
q.push(to2[i]);
vis[to2[i]]=true;
}
}
}
}
void Astar()
{
if(dis1[s]>1e9)
{
printf("-1");
return;
}
if(s==t) k++;
int cnt=,last=-;
priority_queue<node>q;
now.num=s;
now.dis=;
q.push(now);
while(!q.empty())
{
now=q.top();
q.pop();
if(now.num==t)
{
cnt++;
if(cnt==k)
{
printf("%d",now.dis);
return;
}
}
for(int i=front[now.num];i;i=nxt[i])
{
nt.num=to[i];
nt.dis=now.dis+val[i];
q.push(nt);
}
}
printf("-1");
}
int main()
{
init();
spfa();
Astar();
}
最新文章
- 设计模式--建造者模式Builder(创建型)
- PHP+Nginx环境搭配
- CSS3弹性伸缩布局(一)——box布局
- 【转】浅析linux内存模型
- C#将HTML导出Excel
- Python学习笔记 第二课 循环
- javascript设计模式5
- C# .NET 获取枚举值的自定义属性(特性/注释/备注)信息
- 关于mtk Android打开串口权限问题
- iOS 圆角那些事(转)
- 【翻译】使用Visual Studio在Azure上部署Asp.Net Core Web应用
- open_links_per_instance 和 open_links 参数说明
- spring cloud 创建一个简单Eureka Server
- Nodejs通过账号密码连接MongoDB数据库
- L0 Regularization
- Javascript异步编程的4种方法(阮一峰)
- 尚硅谷redis学习4-数据类型
- include方便查找
- 团体程序设计天梯赛L3-019 代码排版(23分)
- Mac OS Yosemite 文件批量重命名