In Action

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6663 Accepted Submission(s): 2265

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

Sample Output

5
impossible

每辆坦克只能占领一个据点，求出从出发点到每个据点的最短路，然后就就是每个据点占领或不占领的选择，即转化为01背包问题

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
int d[102], INF = 1e7, pow1[102], dp[1000002], sum;
using namespace std; //d[n]表示从出发点到n的最短路，dp[n]为耗油量n时所能占领的最大pow
struct node
{
    int begin, end, cost;
}go[10003];
void init(int n)
{
    for (int i = 0; i <= n; i++)
        d[i] = INF;
    d[0] = 0;
}
void ff(int m, int n)
{
    bool bb = 1;
    while (bb)
    {
        bb = 0;
        for (int i = 1; i <= m; i++)
        {
            node e = go[i];
            if (d[e.begin] != INF && d[e.end]>d[e.begin] + e.cost)
                d[e.end] = d[e.begin] + e.cost, bb = 1;
            if (d[e.end] != INF && d[e.begin]>d[e.end] + e.cost)
                d[e.begin] = d[e.end] + e.cost;
        }
    }
    sum = 0;
    for (int i = 1; i <= n; i++)
    {
        if (d[i] == INF) continue;//不能到达该点。
        sum += d[i];
    } //若所有能到达的点都占领，耗油量为sum.
}
void DP(int n, int s)
{
    for (int i = 0; i <= s; i++) dp[i] =-INF;
    dp[0] = 0;
    for (int i = 1; i <= n; i++)
    {
        if (d[i] == INF)continue;//不能到达该点
        for (int j = s; j >= d[i]; j--)
            dp[j] = max(dp[j], dp[j - d[i]] + pow1[i]);
    }
}
int main()
{
    int T, m, n; scanf_s("%d", &T);
    while (T--)
    {
        scanf_s("%d%d", &n, &m);
        init(n); int s = 0;
        for (int i = 1; i <= m; i++)
            scanf_s("%d%d%d", &go[i].begin, &go[i].end, &go[i].cost);
        ff(m, n);
        for (int i = 1; i <= n; i++)
        {
            scanf_s("%d", &pow1[i]);
            s += pow1[i];
        }
        /*if (sum >= INF)
        {
            printf("impossible\n");
            continue;
        }*/
        DP(n, sum);
        bool b2 = 1;
        for (int i = 0; i <= sum; i++)//若sum之内还不能占领一半的pow则impossible
        {
            if (dp[i]>s / 2)
            {
                printf("%d\n", i);
                b2 = 0; break;
            }
        }
        if (b2) printf("impossible\n");
    }
}

HDU 3339 In Action(最短路+01背包）

In Action

相关推荐