Connect the Cities[HDU3371]

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18322 Accepted Submission(s): 4482

Problem Description
In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.

Input
The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

Output
For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input
1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

Sample Output
1

kruskal会超时，要用prim。朴素的prim也是极限时间过的，最好加个堆优化。

#include <stdio.h>
#include <string.h>
using namespace std;
class Prim {
#define Prim_MAXN 505
#define Prim_MAXM 100005
public:
    int N, M;
    int head[Prim_MAXN], dis[Prim_MAXN];
    bool vis[Prim_MAXN];
    struct EDGE {
        int v, d, nex;
    } edge[Prim_MAXM];
    Prim() {
        clear();
    }
    void clear() {
        N = M = ;
        memset(head, -, sizeof(head));
    }
    void addEdge(int a, int b, int c) {
        edge[M].v = b;
        edge[M].d = c;
        edge[M].nex = head[a];
        head[a] = M++;
        edge[M].v = a;
        edge[M].d = c;
        edge[M].nex = head[b];
        head[b] = M++;
    }
    int MST() {
        int ret = , last, next, min;
        for (int i = ; i <= N; i++) {
            dis[i] = 0x7FFFFFFF;
        }
        memset(vis, false, sizeof(vis));
        vis[] = true;
        last = ;
        for (int i = ; i < N; i++) {
            for (int e = head[last]; e != -; e = edge[e].nex) {
                if (dis[edge[e].v] > edge[e].d) {
                    dis[edge[e].v] = edge[e].d;
                }
            }
            min = 0x7FFFFFFF;
            for (int j = ; j <= N; j++) {
                if (dis[j] < min && !vis[j]) {
                    min = dis[j];
                    next = j;
                }
            }
            if (min == 0x7FFFFFFF) {
                return -;
            }
            vis[next] = true;
            ret += dis[next];
            last = next;
        }
        return ret;
    }
};
Prim Pr;
int main() {
    int n, m, t, k;
    scanf("%d", &t);
    while (t--) {
        Pr.clear();
        scanf("%d%d%d", &n, &m, &k);
        Pr.N = n;
        for (int i = ; i < m; i++) {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            Pr.addEdge(a, b, c);
        }
        int nn, x[];
        for (int i = ; i < k; i++) {
            scanf("%d", &nn);
            for (int j = ; j < nn; j++) {
                scanf("%d", &x[j]);
            }
            for (int j = ; j < nn; j++) {
                Pr.addEdge(x[], x[j], );
            }
        }
        printf("%d\n", Pr.MST());
    }
    return ;
}