In Action

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=3339

Description

Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the number of nuclear weapons have soared across the globe.
Nowadays,the crazy boy in FZU named AekdyCoin possesses some nuclear weapons and wanna destroy our world. Fortunately, our mysterious spy-net has gotten his plan. Now, we need to stop it.
But the arduous task is obviously not easy. First of all, we know that the operating system of the nuclear weapon consists of some connected electric stations, which forms a huge and complex electric network. Every electric station has its power value. To start the nuclear weapon, it must cost half of the electric network's power. So first of all, we need to make more than half of the power diasbled. Our tanks are ready for our action in the base(ID is 0), and we must drive them on the road. As for a electric station, we control them if and only if our tanks stop there. 1 unit distance costs 1 unit oil. And we have enough tanks to use.
Now our commander wants to know the minimal oil cost in this action.

Input

The first line of the input contains a single integer T, specifying the number of testcase in the file.
For each case, first line is the integer n(1<= n<= 100), m(1<= m<= 10000), specifying the number of the stations(the IDs are 1,2,3...n), and the number of the roads between the station(bi-direction).
Then m lines follow, each line is interger st(0<= st<= n), ed(0<= ed<= n), dis(0<= dis<= 100), specifying the start point, end point, and the distance between.
Then n lines follow, each line is a interger pow(1<= pow<= 100), specifying the electric station's power by ID order.

Output

The minimal oil cost in this action.
If not exist print "impossible"(without quotes).

Sample Input

2
2 3
0 2 9
2 1 3
1 0 2
1
3
2 1
2 1 3
1
3

Sample Output

5
impossible

HINT

题意

给你一个图,每个图都有权值,然后让你派出多辆坦克,破环至少占权值一半的城市,派出坦克的代价就是从0点到那些城市的距离

求最少代价

题解:

点数很少,注意有重边

直接跑一发flyod,然后再来个背包dp

然后就好了~

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** int dis[][];
int dis1[];
int power[];
int dp[maxn];
int main()
{
//test;
int t=read();
for(int cas=;cas<=t;cas++)
{
int n=read(),m=read();
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
dis[i][j]=inf;
for(int i=;i<m;i++)
{
int a=read(),b=read(),c=read();
dis[a][b]=dis[b][a]=min(dis[a][b],c);
}
int sum,aim;
sum=aim=;
for(int i=;i<=n;i++)
{
power[i]=read();
aim+=power[i];
}
for(int k=;k<=n;k++)
{
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}
}
}
for(int i=;i<=n;i++)
{
dis1[i]=dis[][i];
if(dis1[i]<inf)
sum+=dis1[i];
}
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++)
for(int j=sum;j>=dis1[i];j--)
dp[j]=max(dp[j],dp[j-dis1[i]]+power[i]); int ans=inf;
for(int i=;i<=sum;i++)
{
if(dp[i]>=(aim/+)&&dp[i]<ans)
{
ans=i;
break;
}
}
if(ans>=inf)
printf("impossible\n");
else
printf("%d\n",ans);
}
}

hdu 3339 In Action 背包+flyod的更多相关文章

  1. HDU 3339 In Action(迪杰斯特拉+01背包)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=3339 In Action Time Limit: 2000/1000 MS (Java/Others) ...

  2. HDU 3339 In Action【最短路+01背包】

    题目链接:[http://acm.hdu.edu.cn/showproblem.php?pid=3339] In Action Time Limit: 2000/1000 MS (Java/Other ...

  3. hdu 3339 In Action (最短路径+01背包)

    In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  4. hdu 3339 In Action(迪杰斯特拉+01背包)

    In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  5. HDU 3339 In Action 最短路+01背包

    题目链接: 题目 In Action Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...

  6. HDU 3339 In Action【最短路+01背包模板/主要是建模看谁是容量、价值】

     Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the n ...

  7. HDU 3339 In Action(最短路+背包)题解

    思路:最短路求出到每个点的最小代价,然后01背包,求出某一代价所能拿到的最大价值,然后搜索最后结果. 代码: #include<cstdio> #include<set> #i ...

  8. hdu 3339 In Action

    http://acm.hdu.edu.cn/showproblem.php?pid=3339 这道题就是dijkstra+01背包,先求一遍最短路,再用01背包求. #include <cstd ...

  9. hdu 2546 典型01背包

    分析:每种菜仅仅可以购买一次,但是低于5元不可消费,求剩余金额的最小值问题..其实也就是最接近5元(>=5)时, 购买还没有买过的蔡中最大值问题,当然还有一些临界情况 1.当余额充足时,可以随意 ...

随机推荐

  1. OpenERP 7.0 中文报表PDF乱码(WindowsXP)

    OpenERP默认安装输出的PDF中文报表都是一些方块: 此问题可以通过oecn_base_fonts模块解决: 更多关于oecn_base_fonts的信息请参考: 1. OpenERPv7.0 中 ...

  2. angularJS+requireJS并集成karma测试实践

    最近在为下一个项目做前端技术选型,Angular是必须要用的(BOSS指定,个人感觉也不错,开发效率会很高).由于需要加载的JS很多,所以打算看看angular和requirejs一起用会怎么样.在g ...

  3. SQL Server 最小化日志操作解析,应用[手稿]

    Sql Server 中数据库在BULK_LOGGED/SIMPLE模式下的一些操作会采用最小化日志的记录方式,以减小tran log落盘日志量从而提高整体性能. 这里我简单介绍下哪些操作在什么样的情 ...

  4. java中的类实现comparable接口 用于排序

    import java.util.Arrays; public class SortApp { public static void main(String[] args) { Student[] s ...

  5. 事务处理: databse jdbc mybatis spring

    事务的认识需要一个相当漫长的流程,慢慢在实践中理解,然后在强化相关理论基础. 数据库中的事务: 传统的本地事务处理都是依靠数据库自身事务处理能力,而事务本身是传统关系型数据库的基石.简单来说事务就是一 ...

  6. 看过的bootstrap书籍(附下载地址)

    http://yun.baidu.com/share/link?shareid=3820784617&uk=1008683945 以下书籍下载地址. <BootStrap入门教程> ...

  7. C# html互转mht

    using System;using System.Runtime.InteropServices;using System.Text;using System.IO;namespace HTMLCo ...

  8. php--opp--2.什么是类,什么是对象,类和对象这间的关系

    类的概念:类是具有相同属性和服务的一组对象的集合.它为属于该类的所有对象提供了统一的抽象描述,其内部包括属性和服务两个主要部分.在面向对象的编程语言中,类是一个独立的程序单位,它应该有一个类名并包括属 ...

  9. c++builder CryptoAPI md5

    #include <wincrypt.h> DWORD GetHash( CONST BYTE * pbData, DWORD dwDataLen, ALG_ID algId, LPTST ...

  10. shell's glob

    [shell's glob] basic glob example: range glob example: 参考: http://bash.cumulonim.biz/glob.html