Codeforces Gym 100269D Dwarf Tower spfa
Dwarf Tower
题目连接:
http://codeforces.com/gym/100269/attachments
Description
Little Vasya is playing a new game named “Dwarf Tower”. In this game there are n different items,
which you can put on your dwarf character. Items are numbered from 1 to n. Vasya wants to get the
item with number 1.
There are two ways to obtain an item:
• You can buy an item. The i-th item costs ci money.
• You can craft an item. This game supports only m types of crafting. To craft an item, you give
two particular different items and get another one as a result.
Help Vasya to spend the least amount of money to get the item number 1.
Input
The first line of input contains two integers n and m (1 ≤ n ≤ 10 000; 0 ≤ m ≤ 100 000) — the number
of different items and the number of crafting types.
The second line contains n integers ci — values of the items (0 ≤ ci ≤ 109
).
The following m lines describe crafting types, each line contains three distinct integers ai
, xi
, yi — ai
is
the item that can be crafted from items xi and yi (1 ≤ ai
, xi
, yi ≤ n; ai ̸= xi
; xi ̸= yi
; yi ̸= ai).
Output
The output should contain a single integer — the least amount of money to spend.
Sample Input
5 3
5 0 1 2 5
5 2 3
4 2 3
1 4 5
Sample Output
2
Hint
题意
有n个物品,每个物品的价格是pi,现在你有m种交换方式,就是ai+bi可以换得一个ci
然后问你最便宜得到第一个物品的价钱是多少
题解:
跑最短路,m就相当于建了m条边。
一开始把所有点都压进队列然后跑spfa就好了
至于中途怎么买过来买过去的我不知道,反正暴力出来就是答案咯
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4+5;
int d[maxn];
struct edge{
int x,y;
edge() {}
edge(int X,int Y):x(X),y(Y){}
};
vector<edge>E[maxn];
int inq[maxn];
int main()
{
freopen("dwarf.in","r",stdin);
freopen("dwarf.out","w",stdout);
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&d[i]);
for(int i=1;i<=m;i++)
{
int x,y,z;scanf("%d%d%d",&x,&y,&z);
E[y].push_back(edge(z,x));
E[z].push_back(edge(y,x));
}
queue<int> Q;
for(int i=1;i<=n;i++)
Q.push(i),inq[i]=1;
while(!Q.empty())
{
int now = Q.front();
Q.pop();
inq[now]=0;
for(int i=0;i<E[now].size();i++)
{
int v = E[now][i].x;
int u = E[now][i].y;
if(d[v]+d[now]<d[u])
{
d[u]=d[now]+d[v];
if(!inq[u])
inq[u]=1,Q.push(u);
}
}
}
cout<<d[1]<<endl;
}
Codeforces Gym 100269D Dwarf Tower spfa的更多相关文章
- Codeforces Gym 100269 Dwarf Tower (最短路)
题目连接: http://codeforces.com/gym/100269/attachments Description Little Vasya is playing a new game na ...
- Problem D. Dwarf Tower spfa
http://codeforces.com/gym/100269/attachments 首先建图,然后图中每条边的权值是会变化的,是由dis[x] + dis[y] ---> dis[m ...
- Codeforces Gym 100269A Arrangement of Contest 水题
Problem A. Arrangement of Contest 题目连接: http://codeforces.com/gym/100269/attachments Description Lit ...
- noip模拟赛 dwarf tower
[问题描述]Vasya在玩一个叫做"Dwarf Tower"的游戏,这个游戏中有n个不同的物品,它们的编号为1到n.现在Vasya想得到编号为1的物品.获得一个物品有两种方式:1. ...
- dwarf tower
dwarf tower(dwarf.cpp/c/pas)[问题描述]Vasya在玩一个叫做"Dwarf Tower"的游戏,这个游戏中有n个不同的物品,它们的编号为1到n.现在Va ...
- Codeforces Gym 101252D&&floyd判圈算法学习笔记
一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈 ...
- Codeforces Gym 101190M Mole Tunnels - 费用流
题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...
- Codeforces Gym 101623A - 动态规划
题目传送门 传送门 题目大意 给定一个长度为$n$的序列,要求划分成最少的段数,然后将这些段排序使得新序列单调不减. 考虑将相邻的相等的数缩成一个数. 假设没有分成了$n$段,考虑最少能够减少多少划分 ...
- 【Codeforces Gym 100725K】Key Insertion
Codeforces Gym 100725K 题意:给定一个初始全0的序列,然后给\(n\)个查询,每一次调用\(Insert(L_i,i)\),其中\(Insert(L,K)\)表示在第L位插入K, ...
随机推荐
- 最简单的基于FFMPEG的图像编码器(YUV编码为JPEG)(转)
原文转自 https://blog.csdn.net/leixiaohua1020/article/details/25346147/ 伴随着毕业论文的完成,这两天终于腾出了空闲,又有时间搞搞FFMP ...
- 從 kernel source code 查出 版本號碼
kernel/Makefile 1 VERSION = 4 2 PATCHLEVEL = 4 3 SUBLEVEL = 21 4 EXTRAVERSION = 5 NAME = Blurry Fish ...
- rdpClient
https://github.com/jean343/RPI-GPU-rdpClient https://github.com/Nullstr1ng/MultiRDPClient.NET https: ...
- C 实现有追求的线程池 探究
引言 线程池很普通的老话题,讨论的很多.深入的不多,也就那些基础库中才能见到这种精妙完备的技巧.而本文随大流 想深入简述一种高效控制性强的一种线程池实现. 先引入一个概念, 惊群. 简单举个例子. 春 ...
- caffe Python API 之InnerProduct
net.fc3 = caffe.layers.InnerProduct(net.pool1, num_output=1024, weight_filler=dict(type='xavier'), b ...
- ASP.NET 163 smtp服务器响应为:User has no permission
1.问题引出 今天在asp.net程序中,利用System.Net.Mail.MailMessage类和网易163免费邮箱服务器发送邮件时出现了如下问题. 2.解决方案 原因很简单,我们在asp.ne ...
- ora11g listener.ora
配置内容方式1: LISTENER = (DESCRIPTION_LIST = (DESCRIPTION = (ADDRESS = (PROTOCOL = IPC) (KEY = EXTPROC152 ...
- Html Css 练习
一. 取消a链接的下划线 <!DOCTYPE html> <html lang="en"> <head> <meta charset=& ...
- http跟https的区别
http: Hypertext transform protocol 超文本传输协议 是一个为了传输超媒体文档(比如html)的应用层协议 是为了web的浏览器跟web的server端的交流而设计的, ...
- 机器学习方法:回归(三):最小角回归Least Angle Regression(LARS),forward stagewise selection
欢迎转载,转载请注明:本文出自Bin的专栏blog.csdn.net/xbinworld. 希望与志同道合的朋友一起交流,我刚刚设立了了一个技术交流QQ群:433250724,欢迎对算法.技术.应用感 ...