hdu 3339 In Action 背包+flyod

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[20];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int dis[110][110];
int dis1[110];
int power[110];
int dp[maxn];
int main()
{
    //test;
    int t=read();
    for(int cas=1;cas<=t;cas++)
    {
        int n=read(),m=read();
        for(int i=0;i<=n;i++)
            for(int j=0;j<=n;j++)
                dis[i][j]=inf;
        for(int i=0;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=0;
        for(int i=1;i<=n;i++)
        {
            power[i]=read();
            aim+=power[i];
        }
        for(int k=0;k<=n;k++)
        {
            for(int i=0;i<=n;i++)
            {
                for(int j=0;j<=n;j++)
                {
                    dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            dis1[i]=dis[0][i];
            if(dis1[i]<inf)
                sum+=dis1[i];
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;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=0;i<=sum;i++)
        {
            if(dp[i]>=(aim/2+1)&&dp[i]<ans)
            {
                ans=i;
                break;
            }
        }
        if(ans>=inf)
            printf("impossible
");
        else
            printf("%d
",ans);
    }
}