HDU 3339 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): 5220    Accepted Submission(s): 1745

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

 

题意：坦克从0点出发，去破坏每一个电站，每个坦克只能破坏一个电站，每个电站都有一定的电量，问至少破坏1/2的电量最少需要消耗多少油量。

题解：先求出从0点出发到每一个顶点的最短距离，然后把从0出发到每一个顶点的距离之和当作容量v，进行01背包。最后和电量的总和的1/2比较。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;
const int maxn = ;
const int INF = 0x3f3f3f3f;
struct edge
{
    int to,cost;
    friend bool operator < (edge A,edge B)
    {
        return A.cost>B.cost;
    }
};
int n,m;
int dp[maxn*maxn];
vector<edge> g[maxn];
bool done[maxn];
int d[maxn];
int v[maxn];
void dijkstra()
{
    memset(done,false,sizeof(done));
    memset(d,INF,sizeof(d));
    d[] = ;
    priority_queue<edge> q;
    q.push((edge){,});
    while(!q.empty())
    {
        edge cur = q.top();
        q.pop();
        int v = cur.to;
        if(done[v]) continue;
        done[v] = true;
        for(int i = ; i<g[v].size(); i++)
        {
            cur = g[v][i];
            if(d[cur.to]>d[v]+cur.cost)
            {
                d[cur.to] = d[v]+cur.cost;
                q.push((edge){cur.to,d[cur.to]});
            }
        }
    }
}
void solve()
{
    scanf("%d %d",&n,&m);
    int a,b,c;
    for(int i = ; i<=m; i++)
    {
        scanf("%d %d %d",&a,&b,&c);
        g[a].push_back((edge){b,c});
        g[b].push_back((edge){a,c});
    }
    dijkstra();
    double sum = ;
    int count = ;
    memset(dp,,sizeof(dp));
    for(int i = ; i<=n; i++)
    {
        scanf("%d",&v[i]);
        sum += v[i];
        if(d[i]!=INF)
        count += d[i];
    }
    for(int i = ; i<=n; i++)
    {
        for(int j = count; j>=d[i]; j--)
        {
            if(dp[j-d[i]]+v[i]>dp[j])
            {
                dp[j] = dp[j-d[i]]+v[i];
            }
        }
    }
    int flag = ;
    for(int i = ; i<=count; i++)
    {
        if(dp[i]>sum/2.0)
        {
            printf("%d\n",i);
            flag = ;
            break;
        }
    }
    if(flag == ) printf("impossible\n");
    for(int i = ; i<=n; i++) g[i].clear();
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        solve();
    }
    return ;
}