UVa 11374 - Airport Express ( dijkstra预处理 )
起点和终点各做一次单源最短路, d1[i], d2[i]分别代表起点到i点的最短路和终点到i点的最短路,枚举商业线车票cost(a, b); ans = min( d1[a] + cost(a, b) + d2[b] );
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <queue> using namespace std; const int MAXN = ;
const int INF = << ; struct HeapNode
{
int d, u;
HeapNode() { }
HeapNode( int _d, int _u ): d(_d), u(_u) { }
bool operator<( const HeapNode& rhs ) const
{
return d > rhs.d;
}
}; struct Edge
{
int from, to, dist;
Edge() { }
Edge( int f, int t, int d ) : from(f), to(t), dist(d) { }
}; int N, S, E; struct Dijkstra
{
int n, m;
vector<Edge> edges;
vector<int> G[MAXN];
bool done[MAXN];
int d1[MAXN];
int d2[MAXN];
int p1[MAXN];
int p2[MAXN]; void init( int n )
{
this->n = n;
for ( int i = ; i <= n; ++i ) G[i].clear();
edges.clear();
return;
} void AddEdge( int from, int to, int dist )
{
edges.push_back( Edge( from, to, dist ) );
m = edges.size();
G[from].push_back(m - );
return;
} void dijkstra( int s, int *d, int *p )
{
priority_queue<HeapNode> Q;
for ( int i = ; i <= n; ++i ) d[i] = INF;
d[s] = ;
memset( done, , sizeof(done) );
Q.push( HeapNode( , s ) );
while ( !Q.empty() )
{
HeapNode x = Q.top();
Q.pop();
int u = x.u;
if ( done[u] ) continue;
done[u] = true;
for ( int i = ; i < G[u].size(); ++i )
{
Edge& e = edges[ G[u][i] ];
if ( d[e.to] > d[u] + e.dist )
{
d[e.to] = d[u] + e.dist;
p[e.to] = G[u][i];
Q.push( HeapNode( d[e.to], e.to ) );
}
}
}
return;
} void Print1( int e )
{
if ( e == S )
{
printf( "%d", S );
return;
}
Edge& tp = edges[ p1[e] ];
Print1( tp.from );
printf( " %d", e );
return;
} void Print2( int e )
{
if ( e == E )
{
printf( " %d", E );
return;
}
Edge& tp = edges[ p2[e] ];
printf( " %d", e );
Print2( tp.from );
return;
} }; int M, K;
Dijkstra slv;
vector<Edge> ec; //经济线路 int main()
{
//freopen( "UVa_11374.in", "r", stdin );
//freopen( "UVa_11374.out", "w", stdout );
int cas = ;
while ( scanf( "%d%d%d", &N, &S, &E ) == )
{
slv.init( N );
scanf( "%d", &M );
for ( int i = ; i < M; ++i )
{
int u, v, cost;
scanf( "%d%d%d", &u, &v, &cost );
slv.AddEdge( u, v, cost );
slv.AddEdge( v, u, cost );
} ec.clear();
scanf( "%d", &K );
for ( int i = ; i < K; ++i )
{
int u, v, cost;
scanf( "%d%d%d", &u, &v, &cost );
ec.push_back( Edge(u, v, cost) );
} slv.dijkstra( S, slv.d1, slv.p1 );
slv.dijkstra( E, slv.d2, slv.p2 ); // for ( int i = 1; i <= N; ++i )
// {
// printf("%d %d\n", slv.d1[i], slv.d2[i] );
// } int ans = slv.d1[E];
int huan = -;
int huaned; for ( int i = ; i < K; ++i )
{
int u = ec[i].from;
int v = ec[i].to;
int tmp1 = slv.d1[u] + ec[i].dist + slv.d2[v];
int tmp2 = slv.d1[v] + ec[i].dist + slv.d2[u];
if ( tmp1 < ans )
{
ans = tmp1;
huan = u;
huaned = v;
}
if ( tmp2 < ans )
{
ans = tmp2;
huan = v;
huaned = u;
}
} if (cas) puts("");
++cas; if ( huan == - )
{
slv.Print1(E);
puts("");
puts("Ticket Not Used");
}
else
{
slv.Print1(huan);
slv.Print2(huaned);
puts("");
printf( "%d\n", huan );
}
printf( "%d\n", ans ); }
return ;
}
UVa 11374 - Airport Express ( dijkstra预处理 )的更多相关文章
- UVA - 11374 Airport Express (Dijkstra模板+枚举)
Description Problem D: Airport Express In a small city called Iokh, a train service, Airport-Express ...
- UVA - 11374 - Airport Express(堆优化Dijkstra)
Problem UVA - 11374 - Airport Express Time Limit: 1000 mSec Problem Description In a small city c ...
- UVA 11374 Airport Express SPFA||dijkstra
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&p ...
- UVA 11374 Airport Express 机场快线(单源最短路,dijkstra,变形)
题意: 给一幅图,要从s点要到e点,图中有两种无向边分别在两个集合中,第一个集合是可以无限次使用的,第二个集合中的边只能挑1条.问如何使距离最短?输出路径,用了第二个集合中的哪条边,最短距离. 思路: ...
- UVA 11374 Airport Express (最短路)
题目只有一条路径会发生改变. 常见的思路,预处理出S和T的两个单源最短路,然后枚举商业线,商业线两端一定是选择到s和t的最短路. 路径输出可以在求最短路的同时保存pa数组得到一棵最短路树,也可以用di ...
- UVA 11374 Airport Express(最短路)
最短路. 把题目抽象一下:已知一张图,边上的权值表示长度.现在又有一些边,只能从其中选一条加入原图,使起点->终点的距离最小. 当加上一条边a->b,如果这条边更新了最短路,那么起点st- ...
- UVA 11374 Airport Express(枚举+最短路)
枚举每条商业线<a, b>,设d[i]为起始点到每点的最短路,g[i]为终点到每点的最短路,ans便是min{d[a] + t[a, b] + g[b]}.注意下判断是否需要经过商业线.输 ...
- uva 11374 最短路+记录路径 dijkstra最短路模板
UVA - 11374 Airport Express Time Limit:1000MS Memory Limit:Unknown 64bit IO Format:%lld & %l ...
- 训练指南 UVA - 11374(最短路Dijkstra + 记录路径 + 模板)
layout: post title: 训练指南 UVA - 11374(最短路Dijkstra + 记录路径 + 模板) author: "luowentaoaa" catalo ...
随机推荐
- Java 截屏工具类
PrintScreenUtils.java package javax.utils; import java.awt.AWTException; import java.awt.Dimension; ...
- Getting aCC Error :name followed by "::" must be a class or namespace name"
Getting aCC Error :name followed by "::" must be a class or namespace name" 原始是这样子的: ...
- 旧文备份:Python国际化支持
Python通过gettext模块支持国际化(i18n),可以实现程序的多语言界面的支持,下面是我的多语言支持实现: 在python安装目录下的./Tools/i18n/(windows下例 D:\P ...
- SpringBoot非官方教程 | 第二十五篇:2小时学会springboot
转载请标明出处: http://blog.csdn.net/forezp/article/details/61472783 本文出自方志朋的博客 一.什么是spring boot Takes an o ...
- Python 初始—(字符编码解码)
字符编码之间的编码转换则需要通过Unicode 进行转换,那么需要一个编码和解码实现与Unicode进行关联转换 例如utf-8转gbk utf-8----decode----->Unicode ...
- 爬虫学习(十四)——xpath项目实践
import osimport timeimport urllib.requestimport urllib.parsefrom lxml import etree # 构建面向对象的代码方式clas ...
- udp发送广播消息
import socket if __name__ == '__main__': # 创建udpsocket udp_socket = socket.socket(socket.AF_INET, so ...
- VM虚拟机里的Ubuntu系统怎么设置屏幕分辨率
说白了就是安装VMWare tools工具,步骤如下: 1)在VMWare中启动ubuntu虚拟机 2)在VMWare中:右键单击启动虚拟机,点击[安装vmware tools] 3)在ubuntu中 ...
- DrawGrid 做图片显示 代码简单 参考性强 (Delphi7)
运行效果图 源码 http://files.cnblogs.com/lwm8246/DrawGrid_demo.rar procedure TfrmMain.GridDrawCell(Send ...
- Python常用函数记录
Python常用函数/方法记录 一. Python的random模块: 导入模块: import random 1. random()方法: 如上如可知该函数返回一个[0,1)(左闭右开)的一个随机的 ...