6.2.8 In Action

In Action

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

Total Submission(s): 285    Accepted Submission(s): 121

 

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

22 30 2 92 1 31 0 2132 12 1 313

Sample Output

5impossible

最短路径+01背包

#include<cstdio>#include<cstring>#include<utility>#include<queue>#define INF 1e8using namespace std;typedef pair<int, int> pii;const int MAXN = 100 + 10;const int MAXM = 2 * 10000 + 10;int d[MAXN], done[MAXN], first[MAXN];  //dijkstra辅助数组int u[MAXM], v[MAXM], w[MAXM], next[MAXM];int p[MAXN], f[10001];//p记录每个电站的电量 ,f背包问题的辅助数组 .（刚开始f数组开小了，跪好久） int n, m;void init(){for(int i = 0; i <= n; i++){first[i] = -1;done[i] = 0;d[i] = INF;}}void dijkstra(int st){d[st] = 0;priority_queue<pii, vector<pii>, greater<pii> > q;q.push(make_pair(d[st], st));while(!q.empty()){pii p = q.top();q.pop();int u = p.second;if(done[u]) continue;done[u] = 1;for(int e = first[u]; e != -1; e = next[e]){if(d[v[e]] > d[u] + w[e]){d[v[e]] = d[u] + w[e];q.push(make_pair(d[v[e]], v[e]));}}}} int main(){int t;scanf("%d", &t);while(t--){scanf("%d%d", &n, &m);int i, j, a, b, c, index = 0;init();for(i = 0; i < m; i++){scanf("%d%d%d", &a, &b, &c);u[index] = a;v[index] = b;w[index] = c;next[index] = first[u[index]];first[u[index]] = index;index++;u[index] = b;v[index] = a;w[index] = c;next[index] = first[u[index]];first[u[index]] = index;index++;}int sump = 0, sumo = 0;int flag = 0; //是否能够摧毁，0表示不能 for(i = 1; i <= n; i++){scanf("%d", &p[i]);sump += p[i];}sump = (sump >> 1) + 1;dijkstra(0);for(int i = 1; i <= n; i++) {if(d[i] < INF)sumo += d[i];}//ZeroOnePackmemset(f, 0, sizeof(f));for(i = 1; i <= n; i++){for(j = sumo; j >= d[i]; j--){f[j] = max(f[j], f[j-d[i]] + p[i]);}}for(i = 1; i <= sumo; i++){if(f[i] >= sump){flag = 1;break;}}if(flag) printf("%d\n", i);else printf("impossible\n");}return 1;}