In Action（最短路+01背包）

In Action

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 370 Accepted Submission(s): 160

Problem 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

 

Author

Lost@HDU

 

Source

HDOJ Monthly Contest – 2010.03.06

 

Recommend

lcy

 

/*
用蒂克丝特拉算法先算出来从0到n的最小距离，然后利用最小距离和取得的价值进行背包
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 101
#define M 2000005
#define INF 100000000
using namespace std;
int dp[M];
int mapn[N][N];
int v[N];//表示的获得的价值
int dis[N];//表示耗费的路径
int vis[N];
int t,n,m,x,y,val,s;
void dijkstra(int s)
{
    int i,j,minn,pos;
    memset(vis,0,sizeof(vis));
    for(i = 0; i<=n; i++)
        dis[i] = mapn[s][i];
    for(i = 0; i<=n; i++)
    {
        minn = INF;
        for(j = 0; j<=n; j++)
        {
            if(dis[j]<minn && !vis[j])
            {
                pos = j;
                minn = dis[j];
            }
        }
        vis[pos] = 1;
        for(j = 0; j<=n; j++)
        {
            if(dis[pos]+mapn[pos][j]<dis[j] && !vis[j])
                dis[j] = dis[pos]+mapn[pos][j];
        }
    }
}

int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++){
            for(int j=0;j<=n;j++){
                mapn[i][j]=INF;
            }
            mapn[i][i]=0;
        }
        while(m--){
            scanf("%d%d%d",&x,&y,&val);
            if(mapn[x][y]>val){
                mapn[x][y]=val;
                mapn[y][x]=val;
            }
        }
        s=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&v[i]);
            s+=v[i];
        }
        dijkstra(0);//最短路将
        /*
        到目前为止dis是0到各点的距离，v是各个点炸各个点能获得的能量
        */
        /*
        现在求，至少获得一半能量，所需要的走的最短路程
        */
        /*
        dp获得i能量走的最小路径
        */
        //for(int i=1;i<=n;i++)
        //    cout<<dis[i]<<" ";
        //cout<<endl;
//      ok
        //for(int i=1;i<=n;i++)
        //    cout<<v[i]<<" ";
        //cout<<endl;
//      ok
        for(int i=0;i<=s;i++)
            dp[i]=INF;
        dp[0]=0;//初始化
        for(int i=1;i<=n;i++){
            for(int j=s;j>=v[i];j--){
                dp[j]=min(dp[j],dp[j-v[i]]+dis[i]);
            }
        }
        int i;
        //for(i=0;i<=s;i++)
        //    cout<<dp[i]<<" ";
        //cout<<endl;
        int x=s/2+1;
        int minn=INF;
        for(i=x;i<=s;i++){
            if(dp[i]<minn)
                minn=dp[i];
        }
        if(minn==INF)
            printf("impossible\n");
        else
            printf("%d\n",minn);
    }
    return 0;
}