POJ——T 2449 Remmarguts' Date
http://poj.org/problem?id=2449
Time Limit: 4000MS | Memory Limit: 65536K | |
Total Submissions: 30754 | Accepted: 8394 |
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
Source
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue> using namespace std; const int INF(0x3f3f3f3f);
const int M();
const int N();
int n,m,s,t,k;
int sumedge,had[N],hed[N];
struct Edge
{
int v,next,w;
}edge1[M<<],edge2[M<<];
inline void ins(int u,int v,int w)
{
edge1[++sumedge].v=v;
edge1[sumedge].next=hed[u];
edge1[sumedge].w=w;
hed[u]=sumedge;
edge2[sumedge].v=u;
edge2[sumedge].next=had[v];
edge2[sumedge].w=w;
had[v]=sumedge; } int dis[N];
bool inq[N];
void SPFA(int s)
{
for(int i=;i<=n;i++) dis[i]=INF;
queue<int>que; que.push(s);
inq[s]=true; dis[s]=;
for(;!que.empty();)
{
int u=que.front();
que.pop(); inq[u]=;
for(int v,i=had[u];i;i=edge2[i].next)
{
v=edge2[i].v;
if(dis[v]>dis[u]+edge2[i].w)
{
dis[v]=dis[u]+edge2[i].w;
if(!inq[v]) que.push(v),inq[v]=;
}
}
}
} struct Node
{
int g,f,to;
bool operator < (const Node x) const
{
if(f==x.f) return g>x.g;
return f>x.f;
}
};
int Astar()
{
priority_queue<Node>que;
if(dis[s]==INF) return -;
int cnt=; Node now,v;
now.f=dis[s];now.g=;now.to=s;
que.push(now);
for(;!que.empty();)
{
now=que.top(); que.pop();
if(now.to==t) cnt++;
if(cnt==k) return now.g;
for(int i=hed[now.to];i;i=edge1[i].next)
{
v.to=edge1[i].v;
v.g=now.g+edge1[i].w;
v.f=v.g+dis[edge1[i].v];
que.push(v);
}
}
return -;
} int main()
{
scanf("%d%d",&n,&m);
for(int u,v,w;m--;)
{
scanf("%d%d%d",&u,&v,&w);
ins(u,v,w);
}
scanf("%d%d%d",&s,&t,&k);
if(s==t) k++; SPFA(t);
printf("%d\n",Astar());
return ;
}
POJ——T 2449 Remmarguts' Date的更多相关文章
- 【POJ】2449 Remmarguts' Date(k短路)
http://poj.org/problem?id=2449 不会.. 百度学习.. 恩. k短路不难理解的. 结合了a_star的思想.每动一次进行一次估价,然后找最小的(此时的最短路)然后累计到k ...
- 【POJ】2449.Remmarguts' Date(K短路 n log n + k log k + m算法,非A*,论文算法)
题解 (搬运一个原来博客的论文题) 抱着板题的心情去,结果有大坑 就是S == T的时候也一定要走,++K 我发现按照论文写得\(O(n \log n + m + k \ log k)\)算法没有玄学 ...
- poj 2449 Remmarguts' Date(第K短路问题 Dijkstra+A*)
http://poj.org/problem?id=2449 Remmarguts' Date Time Limit: 4000MS Memory Limit: 65536K Total Subm ...
- poj 2449 Remmarguts' Date (k短路模板)
Remmarguts' Date http://poj.org/problem?id=2449 Time Limit: 4000MS Memory Limit: 65536K Total Subm ...
- 图论(A*算法,K短路) :POJ 2449 Remmarguts' Date
Remmarguts' Date Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 25216 Accepted: 6882 ...
- poj 2449 Remmarguts' Date 第k短路 (最短路变形)
Remmarguts' Date Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 33606 Accepted: 9116 ...
- POJ 2449 Remmarguts' Date (第k短路径)
Remmarguts' Date Time Limit: 4000MS Memory Limit: 65536K Total Submissions:35025 Accepted: 9467 ...
- poj 2449 Remmarguts' Date K短路+A*
题目链接:http://poj.org/problem?id=2449 "Good man never makes girls wait or breaks an appointment!& ...
- POJ 2449 - Remmarguts' Date - [第k短路模板题][优先队列BFS]
题目链接:http://poj.org/problem?id=2449 Time Limit: 4000MS Memory Limit: 65536K Description "Good m ...
随机推荐
- 配置 centos apache 的日志文件为每天保存,在home分区
/usr/local/apache/bin/rotatelogs 这个执行程序会根据安装方式不同的位置也不同,yum安装的话,路径为:/usr/sbin/rotatelogs 改为: ErrorLog ...
- POJ 2481 Cows【树状数组】
题意:给出n头牛的s,e 如果有两头牛,现在si <= sj && ei >= ej 那么称牛i比牛j强壮 然后问每头牛都有几头牛比它强壮 先按照s从小到大排序,然后用e来 ...
- 妙用$.extend
在js中,我们有时需要复制一个对象的值,而不是复制它的引用的时候,可以使用jquery的$.extend方法,简单代码如下 <script> var a = { "name&qu ...
- Msql免安装版安装
文首提要: 我下载的MySQL版本是:mysql-5.7.17-winx64.zip Archive版:系统:Windows7 64位. 一.解压文件 下载好MySQL后, ...
- worldcount
码云项目地址 PSP2.1 PSP阶段 预估耗时 (分钟) 实际耗时 (分钟) Planning 计划 20 20 Estimate 估计这个任务需要多少时间 60 70 Development 开发 ...
- 洛谷P4894 GodFly求解法向量
如果没有学过向量相关知识请出门右转高中数学必修四~~~ 当然如果你和我一样也是小学生我也不反对 首先说结论:\(\vec{z}=(y1z2-y2z1,z1x2-z2x1,x1y2-x2y1)\) 其实 ...
- 实现图像剪裁 jquery.Jcrop
配合 jquery.Jcrop 实现上传图片进行剪裁保存功能 <script src="js/jquery.min.js"></script> ...
- 你的Android应用完全不需要那么多的权限
Android系统的权限从用户的角度来看有时候的确有点让人摸不着头脑.有时候可能你只需要做一些简单的事情(对联系人的信息进行编辑),却申请了远超你应用所需的权限(比如访问所有联系人信息的权限). 这很 ...
- Hdu4786
Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...
- WIZnet相关产品介绍
WIZnet 自1998年在韩国创立以来,一致专注研发全硬件TCP/IP协议栈芯片.同一时候开发设计相关网络模块和无线产品,同一时候 WIZnet 鼓舞开源硬件.相关开源硬件产品也已层出不断. 主要 ...