Codeforces Round #287 (Div. 2) E. Breaking Good 路径记录!!!+最短路+堆优化
2 seconds
256 megabytes
standard input
standard output
Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that is really difficult even for experienced gamers.
Walter William, the main character of the game, wants to join a gang called Los Hermanos (The Brothers). The gang controls the whole country which consists of n cities with mbidirectional roads connecting them. There is no road is connecting a city to itself and for any two cities there is at most one road between them. The country is connected, in the other words, it is possible to reach any city from any other city using the given roads.
The roads aren't all working. There are some roads which need some more work to be performed to be completely functioning.
The gang is going to rob a bank! The bank is located in city 1. As usual, the hardest part is to escape to their headquarters where the police can't get them. The gang's headquarters is in city n. To gain the gang's trust, Walter is in charge of this operation, so he came up with a smart plan.
First of all the path which they are going to use on their way back from city 1 to their headquarters n must be as short as possible, since it is important to finish operation as fast as possible.
Then, gang has to blow up all other roads in country that don't lay on this path, in order to prevent any police reinforcements. In case of non-working road, they don't have to blow up it as it is already malfunctional.
If the chosen path has some roads that doesn't work they'll have to repair those roads before the operation.
Walter discovered that there was a lot of paths that satisfied the condition of being shortest possible so he decided to choose among them a path that minimizes the total number of affected roads (both roads that have to be blown up and roads to be repaired).
Can you help Walter complete his task and gain the gang's trust?
The first line of input contains two integers n, m (2 ≤ n ≤ 105, ), the number of cities and number of roads respectively.
In following m lines there are descriptions of roads. Each description consists of three integersx, y, z (1 ≤ x, y ≤ n, ) meaning that there is a road connecting cities number xand y. If z = 1, this road is working, otherwise it is not.
In the first line output one integer k, the minimum possible number of roads affected by gang.
In the following k lines output three integers describing roads that should be affected. Each line should contain three integers x, y, z (1 ≤ x, y ≤ n, ), cities connected by a road and the new state of a road. z = 1 indicates that the road between cities x and y should be repaired and z = 0 means that road should be blown up.
You may output roads in any order. Each affected road should appear exactly once. You may output cities connected by a single road in any order. If you output a road, it's original state should be different from z.
After performing all operations accroding to your plan, there should remain working only roads lying on some certain shortest past between city 1 and n.
If there are multiple optimal answers output any.
2 1
1 2 0
1
1 2 1
4 4
1 2 1
1 3 0
2 3 1
3 4 1
3
1 2 0
1 3 1
2 3 0
8 9
1 2 0
8 3 0
2 3 1
1 4 1
8 7 0
1 5 1
4 6 1
5 7 0
6 8 0
3
2 3 0
1 5 0
6 8 1
In the first test the only path is 1 - 2
In the second test the only shortest path is 1 - 3 - 4
In the third test there are multiple shortest paths but the optimal is 1 - 4 - 6 - 8
给一个图G<V,E>,有N个点和M条边。G的边有两种属性:work or notwork.
现要求选择一条路径Path,满足以下条件:
1. Path起点是1,终点是N
2. Path是所有路经中最短的一条
3. Path里所有notwork的边都要被修复,Path外所有work的边都要被炸毁,且所修复和炸毁的边的总和数最小
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=100000;
int vis[maxn+10],dis[maxn+10],pre[maxn+10],flag[maxn+10],num[maxn+10];
struct Edge{
int to,c,flag,u,v;
}e[maxn+10]; struct node{
int v,dis;
bool operator<(const node a) const{
return this->dis>a.dis;
}
}; vector<Edge> G[maxn+10]; void init(int n)
{
for(int i=1;i<=n;i++) G[i].clear();
MM(vis,0);
MM(dis,inf);
MM(pre,0);
MM(flag,0);
MM(num,0);
} int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
init(n);
for(int i=1;i<=m;i++)
{
scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].flag);
G[e[i].u].push_back((Edge){e[i].v,1,e[i].flag,e[i].u,e[i].v});
G[e[i].v].push_back((Edge){e[i].u,1,e[i].flag,e[i].u,e[i].v});
} priority_queue<node> q;
q.push((node){1,0});
vis[1]=1;
dis[1]=0;
while(q.size())
{
node cur=q.top();q.pop();
int u=cur.v;
vis[u]=1;
if(dis[u]<cur.dis) continue;
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i].to;
if(vis[v]) continue;
if(dis[v]>dis[u]+1)
{
dis[v]=dis[u]+1;
q.push((node){v,dis[v]});
pre[v]=u;
num[v]=num[u]+G[u][i].flag;
}//优先满足最短路
else if(dis[v]==dis[u]+1&&num[v]<num[u]+G[u][i].flag)
{
pre[v]=u;
num[v]=num[u]+G[u][i].flag;
}
}
} int cnt=0;
for(int i=n;i>=1;i=pre[i])
flag[i]=1; for(int i=1;i<=m;i++)
if(flag[e[i].u]&&flag[e[i].v])
{if(!e[i].flag) cnt++;}
else if(e[i].flag) cnt++; printf("%d\n",cnt);
for(int i=1;i<=m;i++)
if(flag[e[i].u]&&flag[e[i].v])
{if(!e[i].flag) printf("%d %d 1\n",e[i].u,e[i].v);}
else if(e[i].flag) printf("%d %d 0\n",e[i].u,e[i].v); }
return 0;
}
分析:这道题主要挂在怎么记录最短路上;
处理方法是给每个节点设置一个pre值,指向在符合要求的最短路上当前节点指向的上一节点,最后
用个flag数组保存路上节点就好,num[i]表示从起点到达i点的最短路上(因为需要先满足最短路)最大
的权值和
Codeforces Round #287 (Div. 2) E. Breaking Good 路径记录!!!+最短路+堆优化的更多相关文章
- Codeforces Round #287 (Div. 2) E. Breaking Good 最短路
题目链接: http://codeforces.com/problemset/problem/507/E E. Breaking Good time limit per test2 secondsme ...
- Codeforces Round #287 (Div. 2) E. Breaking Good [Dijkstra 最短路 优先队列]
传送门 E. Breaking Good time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- 贪心 Codeforces Round #287 (Div. 2) A. Amr and Music
题目传送门 /* 贪心水题 */ #include <cstdio> #include <algorithm> #include <iostream> #inclu ...
- Codeforces Round #287 (Div. 2) C. Guess Your Way Out! 思路
C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standa ...
- CodeForces Round #287 Div.2
A. Amr and Music (贪心) 水题,没能秒切,略尴尬. #include <cstdio> #include <algorithm> using namespac ...
- Codeforces Round #287 (Div. 2) C. Guess Your Way Out! 水题
C. Guess Your Way Out! time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #287 (Div. 2) B. Amr and Pins 水题
B. Amr and Pins time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- Codeforces Round #287 (Div. 2) A. Amr and Music 水题
A. Amr and Music time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Codeforces Round #287 (Div. 2) D. The Maths Lecture [数位dp]
传送门 D. The Maths Lecture time limit per test 1 second memory limit per test 256 megabytes input stan ...
随机推荐
- 【VS开发】【图像处理】ISP图像传感器处理器基础
1 前言 做为拍照手机的核心模块之一,camera sensor效果的调整,涉及到众多的参数,如果对基本的光学原理及sensor软/硬件对图像处理的原理能有深入的理解和把握的话,对我们的 ...
- [ASP.NET] [JS] GridView点击高亮当前选择行,并在点击另一行时恢复上一选择行背景颜色
在ASP.NET中的gridview控件里面可以通过设定其OnRowDataBound事件来进行实现高亮当前行的操作 前端控件的设置: 只要设置好OnRowDataBound属性即可,会自动在.cs文 ...
- GTS原理、架构
全局事务服务(Global Transaction Service,简称 GTS)是阿里新推出的分布式事务处理方案. 1. GTS 的目标 GTS是一个面向互联网交易场景的分布式事务解决方案. 制约分 ...
- Python常用方法库备忘(一)_当前路径下文件夹和文件
#!/usr/bin/env python # -*- coding:utf-8 -*- # --------------*-------------- # @Author : AilF # @Tim ...
- zookeeper集群的搭建(CentOS 7)
注意ip地址为: 虚拟机ip设置 TYPE="Ethernet"BOOTPROTO="static"NAME="enp0s3"DEVICE= ...
- Select 和Alert
Select 和Alert使用前都必须先导入 from selenium.webdriver.common.alert import Alert from selenium.webdriver.sup ...
- Java获取文件的后缀名。
/** * 详细步骤 */ private static void test1() { //获取文件的原始名称 String originalFilename = "tim.g (1).jp ...
- 利用bing图片搜索接口开发图片搜索应用程序
概述:通过bing的图片搜索引擎,开发自己的图片搜索应用程序.bing的图片搜索接口是收费的,但是初次注册使用,key可以免费试用30天 程序运行效果如下 一,代码如下 static SearchRe ...
- 剑指offer-2:斐波那契数列
二.斐波那契数列 题目描述 大家都知道斐波那契数列,现在要求输入一个整数n,请你输出斐波那契数列的第n项(从0开始,第0项为0). n<=39 1.递归法 1). 分析 斐波那契数列的标准公式为 ...
- MySQL第二讲 一一一一 MySQL语句进阶
通过命令来备份数据库: 通过数据库软件里面的,mysqldump模块来操作,如下: mysqldump -u root db1 > db1.sql -p; //没有-d就是备份的时候:数据表结构 ...