HDU - 3339 In Action

题目：
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

代码如下：

#include<bits/stdc++.h>
using namespace std;
#define MAX 105
#define NIL 0x3f3f3f3f
int n,m,mp[MAX][MAX],dis[MAX],aa[MAX],dp[MAX * 100];
bool vis[MAX];
void Dijkstra()
{
    memset(dis,NIL,sizeof(dis));
    memset(vis,false,sizeof(vis));
    dis[0] = 0;
    priority_queue<int> q;
    q.push(0);
    while(!q.empty()){
        int flag = q.top();
        q.pop();
        vis[flag] = true;
        for(int i = 0;i <= n;i++){
            if(dis[i] > dis[flag] + mp[flag][i]){
                dis[i] = dis[flag] + mp[flag][i];
                q.push(i);
            }
        }
    }
}
int main()
{
    int t,a,b,x,sum,sum2;
    cin >> t;
    while(t--){
        int flag = 0;
        cin >> n >> m;
        sum2 = sum = 0;
        memset(mp,NIL,sizeof(mp));
        for(int i = 0;i < m;i++){
            cin >> a >> b >> x;
            if(mp[a][b] > x) mp[a][b] = mp[b][a] = x;
        }
        Dijkstra();
        for(int i = 1;i <= n;i++){
            cin >> aa[i];
            sum2 += aa[i];//总电量
            if(dis[i] != 0x3f3f3f3f) sum += dis[i];//计算背包容量
        }
        memset(dp,0,sizeof(dp));
        for(int i = 1;i <= n;i++){//0-1背包问题
            for(int j = sum;j >= dis[i];j--){
                dp[j] = max(dp[j],dp[j - dis[i]] + aa[i]);
            }
        }
        for(int i = 1;i <= sum;i++){
            if(dp[i] > sum2 / 2){
                flag = 1;
                cout << i << endl;
                break;
            }
        }
        if(!flag) cout << "impossible" << endl;
    }
    return 0;
}

题意：
输入t组数据，第一行n,m代表有n个发电站以及m条路径，接下来输入m条路径以及这n个发电站的总电量。现在你要摧毁其中的一些发电站，要求摧毁的发电站的电量要大于总电量一半，并且走过的路径要尽可能的短。
思路：
个人觉得这是一道好题，先用Dijkstra算法计算出0点到各个点的最短路径，然后用0 - 1背包，dp[i]代表当摧毁的发电站电量为i时走的最短路径。