http://poj.org/problem?id=2449

Remmarguts' Date
Time Limit: 4000MS   Memory Limit: 65536K
Total Submissions: 18168   Accepted: 4984

Description

"Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks' head, he told them a story. 
"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 first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T. 
The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).

Output

A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output "-1" (without quotes) instead.

Sample Input

2 2
1 2 5
2 1 4
1 2 2

Sample Output

14

Source

POJ Monthly,Zeyuan Zhu
 
【题解】:

给定一个图,求起点到终点的第k短路。

分析:先用dijkstra从t反向寻找最短路。然后使用A*算法,把f(i)=g(i) + h(i)。h(i)就是i点到t的最短距离。当某点出队次数达到k次的时候,结果为该点的当前路程+该点到t的最短距离。(我没有判断不连通的情况)

为什么这样做是对的呢?我们这样来思考,如果不实用最短路,而只使用A*那么t第x次出队的结果即为第x短路的距离。继而可以想到,从第一个出队次数达到x的点,沿着最短路走到t,一定是第x短路。

说实话我也没有完全理解。

另外注意s==t的情况,据说k要++,不明白为啥。

这个题用来测第k短路的模板真的再好不过了,累计我wa在这么几个地方:
1.起点和终点是同一点时一定要走,不能输出0;
2.用普通邻接矩阵的话要注意重边的问题(所以强烈推荐邻接表水所有题);

【code】:

 /**
Judge Status:Accepted Memory:9252K
Time:204MS Language:G++
Code Length:2590B Author:cj
*/ #include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<string.h> #define N 1005
#define M 200100
#define INF 1000000000
using namespace std; struct Edge
{
int v,next,c;
}edge[M]; struct Nod //A*中的优先队列数据类型
{
int v,d,h;
}; struct Dj //dijkstra中的优先队列数据类型
{
int u,dis;
}; bool operator < (Dj a,Dj b) //dijkstra中的优先对队重载<号
{
return b.dis<a.dis;
} bool operator < (Nod a,Nod b) //A*中的优先队列重载<号
{
return b.h+b.d<a.h+a.d;
} int head[N];
int tail[N];
int dis[N];
int visit[N];
int edge_cnt;
int cnt[N]; void init() //初始化
{
memset(head,-,sizeof(head));
memset(tail,-,sizeof(tail));
edge_cnt = ;
memset(visit,,sizeof(visit));
memset(cnt,,sizeof(cnt));
} void addEdge(int a,int b,int c)
{
edge[edge_cnt].v = b;
edge[edge_cnt].c = c;
edge[edge_cnt].next = head[a];
head[a] = edge_cnt++; //记录起始边 edge[edge_cnt].v = a;
edge[edge_cnt].c = c;
edge[edge_cnt].next = tail[b];
tail[b] = edge_cnt++; //记录结束边
}
void Dijkstra(int n,int t) //对于以各点为起点到t点的最短路径
{
priority_queue<Dj> p_q;
Dj now,temp;
int i;
for(i=;i<=n;i++) dis[i]=INF;
dis[t] = ;
temp.dis = ;
temp.u = t;
p_q.push(temp);
while(!p_q.empty())
{
temp = p_q.top();
p_q.pop();
if(visit[temp.u]) continue;
visit[temp.u] = ;
for(i=tail[temp.u];i!=-;i=edge[i].next)
{
int v = edge[i].v;
if(dis[v]>dis[temp.u]+edge[i].c)
{
dis[v]=dis[temp.u]+edge[i].c;
now.u = v;
now.dis = dis[v];
p_q.push(now);
}
}
}
} int Astar_Kth(int s,int t,int k)
{
Nod cur,nxt;
priority_queue<Nod> p_q;
cur.v = s;
cur.d = ;
cur.h = dis[s];
p_q.push(cur);
while(!p_q.empty())
{
cur = p_q.top();
p_q.pop();
cnt[cur.v]++;
if(cnt[cur.v]>k) continue;
if(cnt[t]==k) return cur.d;
int i;
for(i=head[cur.v];i!=-;i=edge[i].next)
{
int v = edge[i].v;
nxt.d = cur.d + edge[i].c;
nxt.v = v;
nxt.h = dis[v];
p_q.push(nxt);
}
}
return -;
} int main()
{
int n,m;
scanf("%d%d",&n,&m);
int i;
int a,b,c;
init();
for(i=;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
addEdge(a,b,c);
}
int s,t,k;
scanf("%d%d%d",&s,&t,&k);
if(s==t) k++;
Dijkstra(n,t);
printf("%d\n",Astar_Kth(s,t,k));
return ;
}

poj 2449 Remmarguts' Date(第K短路问题 Dijkstra+A*)的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. poj 2449 Remmarguts' Date(K短路,A*算法)

    版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/u013081425/article/details/26729375 http://poj.org/ ...

  5. POJ 2449 Remmarguts' Date ( 第 k 短路 && A*算法 )

    题意 : 给出一个有向图.求起点 s 到终点 t 的第 k 短路.不存在则输出 -1 #include<stdio.h> #include<string.h> #include ...

  6. 【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 ...

  7. 【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 ...

  8. 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!& ...

  9. 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 ...

随机推荐

  1. poj 2182 树状数组

    这题对于O(n^2)的算法有很多,我这随便贴一个烂的,跑了375ms. #include<iostream> #include<algorithm> using namespa ...

  2. hdu 1518 深搜

    #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #i ...

  3. 一个简单的XML与数组之间的转换

    xml是网络使用最多的数据交换格式,所以,不掌握怎么操作它,又有蛋疼的了. php中可以操作xml的类/函数很多,个人认为最简单的是SimpleXMLElement这个类,它的使用就跟其名字一样:简单 ...

  4. Linux 命令 - jobs: 显示后台作业的状态信息

    命令格式 jobs [-lnprs] [jobspec ...] jobs -x command [args] 命令参数 -l 额外显示作业的进程 ID. -n 只列出状态发生变化的进程. -p 只列 ...

  5. Android drawable xml 各种小知识

    摘抄自网络. 圆角或者各种变种背景, <?xml version="1.0" encoding="utf-8"?> <shape xmlns: ...

  6. Sqlite事物与锁

    1事务 事务定义了一组SQL命令的边界,这组命令或者作为一个整体被全部执行,或者都不执行.事务的典型实例是转帐. 2事务的范围 事务由3个命令控制:BEGIN.COMMIT和ROLLBACK.BEGI ...

  7. 北大ACM(POJ1001-Exponentiation)

    Question:http://poj.org/problem?id=1001问题点:大数运算(求幂) #include <iostream> using namespace std; # ...

  8. chattr实现文件不可删除

    用自己的话解释清楚这件事儿~ 目前问题: Android手机,在/system/app 目录下的apk,使用chmod 修改权限失败,rm命令也删除不掉. 现象: rm failed for wand ...

  9. 总是你 2008-3 (献给L之一)

    文/安然 总是你 是梦里的那份最温柔 轻轻碰触 不再敢轻易提及 总是你 是不经意的那份最感动 点点鲜活 从来就没有淡去 时间远走 我轻轻的收起 那段记忆 不再开启 但是 偶尔 偶尔 总是 总是在不经意 ...

  10. windbg基本命令

    1, .reload k 当前调用堆栈.u 当前正在执行的代码. 2, ~ 查看被调试进程中的线程信息每一行是一个线程的信息.第一行中,0 表示这个进程的编号:1ff4.1038 是 16 进制数字, ...