mos mean i++ disco string sum 建設 out ssi

題目鏈接：

http://acm.hdu.edu.cn/showproblem.php?pid=3371

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

Author dandelion

Source HDOJ Monthly Contest – 2010.04.04

 1 /*
 2 問題
 3 輸入已經存在的圖以及將要增加的邊及其花費，計算並輸出最小生成樹還需要的最小花費
 
 4 
 5 解題思路
 6 將現在的圖連起來使得它們的花費為0，再加入將要建設的邊，直接跑一邊prim即可，註意不能構成最小生成樹
 7 的情況說出-1，代碼中的u == -1 結束很關鍵。 
 8 */ 
 9 #include<cstdio>
10 #include<cstring>
11 
12 const int N=550;
13 const int INF=99999999;
14 int e[N+10][N+10],a[N],dis[N],bk[N];
15 int prim();
16 int n,m,k;
17 
18 int main(){
19     int T,i,j,q,t1,t2,t3,t;
20     scanf("%d",&T);
21     while(T--){
22         for(i=1;i<=N;i++){
23             for(j=1;j<=N;j++){
24                 e[i][j] = i==j?0:INF;
25             }
26         }
27         scanf("%d%d%d",&n,&m,&k);
28         for(i=1;i<=m;i++){
29             scanf("%d%d%d",&t1,&t2,&t3);
30             if(e[t1][t2] > t3){
31                 e[t1][t2] = t3;
32                 e[t2][t1] = t3;
33             }    
34         }
35         for(i=1;i<=k;i++){
36             scanf("%d",&t);
37             for(j=1;j<=t;j++){
38                 scanf("%d",&a[j]);
39             }
40             for(j=1;j<=t-1;j++){
41                 for(q=j+1;q<=t;q++){
42                     e[ a[j] ][ a[q] ]=0;
43                     e[ a[q] ][ a[j] ]=0;
44                 }
45             }
46         }
47         
48         /*for(i=1;i<=n;i++){
49             for(j=1;j<=n;j++){
50                 printf("%9d",e[i][j]);
51             }
52             printf("\n");
53         }*/
54         int ans=prim();
55         if(ans == -1)
56             printf("-1\n");
57         else
58             printf("%d\n",ans);
59     }
60     return 0;
61 } 
62 
63 int prim()
64 {
65     int i;
66     for(i=1;i<=n;i++)
67         dis[i]=e[1][i];
68     memset(bk,0,sizeof(bk));
69     bk[1]=1;
70     int c=1,sum=0,mina,u;
71     
72     while(c < n){
73         mina=INF;
74         u=-1;
75         for(i=1;i<=n;i++){
76             if(!bk[i] && dis[i] < mina){
77                 mina=dis[i];
78                 u=i;
79             }
80         }
81         //printf("u==%d\n",u);
82         if(u == -1)
83             break;
84         bk[u]=1;
85         c++;
86         sum += dis[u];
87         for(i=1;i<=n;i++){
88             if(!bk[i] && dis[i] > e[u][i])
89                 dis[i] = e[u][i];
90         }
91     }
92     if(u == -1)
93         return -1;
94     return sum;    
95 }

HDU 3371 Connect the Cities（prim算法）