Connect the Cities + prime算法 + (超时)并查集

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6008    Accepted Submission(s): 1747

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

16 4 31 4 22 6 12 3 53 4 332 1 22 1 33 4 5 6

 

Sample Output

1
/*思路：最小生成树算法，将还存在的路费用置为0，需要维修的路费用置为C（这地方注意输入数据的重复性，要取最小的一个值）*/
#include<iostream>#include<cstdio>using namespace std ;#define MAX 505#define inf 1000000000int s[MAX][MAX] , vi[MAX] , n;/*void prime()  //用prime算法计算最小生成树距离（使用这种算法超时）{int i , j , sum = 0 ;while(true){int min1 = inf , k =- 1;for( i = 1 ; i <= n ; i ++){if(vi[i]){for( j = 1 ; j <= n ; j ++)if(!vi[j] && min1 > s[i][j]){k = j ;min1 = s[i][j] ;}}}if( k == - 1)break;vi[k] =  1;sum += min1 ;}int cnt = 0 ;for( i = 1  ; i <= n ; i ++)if(vi[i])cnt ++ ;if(cnt == n && sum < inf )//cout << sum << endl ;printf("%d\n",sum);else//cout << "-1" << endl ;printf("-1\n");return ;}*/void prime(){int i,j,k,min,dist[MAX],sum=0;for(i=1;i<=n;i++)dist[i]=s[1][i];vi[1]=1;for(i=2;i<=n;i++){min=inf;k=i;for(j=1;j<=n;j++)if(!vi[j]&&dist[j]<min){min=dist[j];k=j;}sum=sum+min;if(sum>=inf)break;vi[k]=1;for(j=1;j<=n;j++)if(!vi[j]&&s[k][j]<dist[j])dist[j]=s[k][j];}if(sum>=inf)printf("-1\n");elseprintf("%d\n",sum);}int main(void){int t ;//cin>> t;scanf("%d",&t);while(t --){int  m , k ;//cin>> n >> m >> k ;scanf("%d %d %d",&n , &m , &k);int i , j , l ;memset( vi , 0 , sizeof(vi));for( i = 0 ; i < MAX ; i ++) //初始化各个点的距离，置为无穷大for( j = 0 ; j < MAX ; j ++){s[i][j] = s[j][i] = inf ;if( i == j )s[i][j] = s[j][i] = 0 ;}for( i = 0 ; i < m ; i ++){int A , B , C ;//cin >> A >> B >> C ;scanf("%d %d %d",&A,&B,&C);s[A][B] = s[B][A] = s[A][B] > C ? C : s[A][B] ; //避免重复输入，取最小值}for( i = 0 ; i < k ; i ++) //将还存在的路置为0，即最小花费为0{int  t ,a[MAX];//cin >> t ;scanf("%d",&t);for( j = 0 ; j < t ; j ++)//cin >> a[j] ;scanf("%d",&a[j]);for( j = 0 ; j < t ; j ++)for( l = j + 1 ; l < t ; l ++ )s[a[j]][a[l]] = s[a[l]][a[j]] = 0 ;}vi[1] = 1 ;//从标号为1的点，开始进行prime算法prime();}return 0 ;}
 
/*思路：并查集(超时算法) */
#include<stdio.h>#include<algorithm>using namespace std;#define MAX 505int father[MAX];typedef struct node{int a,b;int c;}NODE;NODE ca[150005];bool cmp(NODE x,NODE y){if(x.c<y.c){return 1;}return 0;}int findset(int x){return x!=father[x]?father[x]=findset(father[x]):x;}int kruskal(int m){int i,mincost=0,x,y;for(i=0;i<m;i++){x=ca[i].a;y=ca[i].b;x=findset(x);y=findset(y);if(x!=y){father[x]=y;mincost+=ca[i].c;}}return mincost;}int main(){int n,m,k,i,mincost=0 , t ;scanf("%d",&t);while( t--){scanf("%d %d %d",&m,&n,&k);for(i=0;i<n;i++){scanf("%d %d %d",&ca[i].a,&ca[i].b,&ca[i].c);}int kk = i ;while(k--){int t1 , a[MAX];scanf("%d",&t1);for( i = 0 ; i < t1 ; i ++ )scanf("%d",&a[i]);for( i = 0 ; i <t1 ; i ++)for( int j = i + 1; j < t1 ; j ++){ca[kk].a = a[i] ;ca[kk].b = a[j] ;ca[kk++].c = 0 ;}}for(i=0;i<m;i++){father[i]=i;}sort(ca,ca+kk,cmp);int l = 1 ;for( i = 1 ; i < kk ; i ++) //去掉重复的路，取最小值{if(ca[i].a == ca[i-1].a && ca[i].b == ca[i-1].b )continue ;ca[l].a = ca[i].a ;ca[l].b = ca[i].b ;ca[l++].c = ca[i].c ;}mincost=kruskal(l);int count=0;for(i=0;i<m;i++){if(father[i]==i){count++;}}if(count>1){puts("?");}else{printf("%d\n",mincost);}}return 0;}