Connect the Cities

Connect the Cities

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

Total Submission(s): 360    Accepted Submission(s): 129

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   C++提交就ac了。G++超时。   #include #include #include using namespace std; const int MAXN = 500 + 10; const int MAXM = 25000 + 10; int p[MAXN], vis[MAXN]; int u[MAXM], v[MAXM], r[MAXM], w[MAXM]; int cmp(int i, int j){ return w[i] < w[j]; } int find_set(int i){ if(i != p[i]) p[i] = find_set(p[i]); return p[i]; } void union_set(int i, int j){ int a = find_set(i); int b = find_set(j); p[b] = a; } int main(){ int t, n, m, k; int a, b, c; scanf("%d", &t); while(t--){ scanf("%d%d%d", &n, &m, &k); memset(vis, 0, sizeof(vis)); for(int i = 1; i <= n; i++) p[i] = i; for(int i = 0; i < m; i++){ scanf("%d%d%d", &a, &b, &c); u[i] = a; v[i] = b; w[i] = c; } for(int i = 0; i < m; i++) r[i] = i; sort(r, r + m, cmp); int len = 0, sum = 0; //sum记录了已加入连通分支中的顶点数目 for(int i = 0; i < k; i++){ scanf("%d", &len); scanf("%d", &a); int x = find_set(a); for(int j = 0; j < len-1; j++){ scanf("%d", &b); int y = find_set(b); if(x != y){ union_set(x, y); if(!vis[a]){ vis[a] = 1; sum++; } if(!vis[b]){ vis[b] = 1; sum++; } } } } int cnt = 0; for(int i = 0; i < m; i++){ int e = r[i]; int x = find_set(u[e]); int y = find_set(v[e]); if(x != y) { if(!vis[u[e]]){ vis[u[e]] = 1; sum++; } if(!vis[v[e]]){ vis[v[e]] = 1; sum++; } union_set(x, y); cnt += w[e]; } } int flag = 1; for(int i = 1; i <= n; i++){ for(int j = i+1; j <= n; j++){ if(find_set(i) != find_set(j)){ flag = 0; break; } } } if(sum == n) printf("%d\n", cnt); else printf("-1\n"); } return 1; }