Connect the CitiesTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11727Accepted Submission(s): 3278
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
Author
dandelion
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FUUUUUUUUUUUUUUUUUUUCK，，我又忘了判断重边！！！！
好吧，是我太笨了！
首先这道题很简单，我们只要两重循环把相连的城市之间的距离设置为0就可以了，
Prim算法400MS+，，而Kruskal算法700MS+，并且我还做了并查集优化。
Prim算法代码：#include <stdio.h>#include <string.h>#define MAX 550#define INF 100000000int graph[MAX][MAX] ;int prim(int n){int sum = 0 ,lowCost[MAX] ;bool visited[MAX] ;memset(visited,false,sizeof(visited)) ;for(int i = 1 ; i <= n ; ++i){lowCost[i] = graph[1][i] ;}visited[1] = true ;for(int i = 1 ; i < n ; ++i){int min = INF , index = -1 ;for(int j = 1 ; j <= n ; ++j){if(!visited[j] && min>lowCost[j]){min = lowCost[j] ;index = j ;}}if(index == -1){if(i < n){return INF ;}break ;}sum += lowCost[index] ;visited[index] = true ;for(int j = 1 ; j <= n ; ++j){if(!visited[j] && lowCost[j]>graph[index][j]){lowCost[j] = graph[index][j] ;}}}return sum ;}int main(){int t ;scanf("%d",&t);while(t–){int n , m , k;scanf("%d%d%d",&n,&m,&k) ;for(int i = 1 ; i <= n ; ++i){graph[i][i] = 0 ;for(int j = 1 ; j < i ; ++j){graph[i][j] = graph[j][i] = INF ;}}for(int i = 0 ; i < m ; ++i){int x, y ,w ;scanf("%d%d%d",&x,&y,&w) ;if(w<graph[x][y]){graph[x][y] = graph[y][x] = w;}}int p[MAX] ;for(int i = 0 ; i < k ; ++i){int t ;scanf("%d",&t);for(int j = 0 ; j < t ; ++j){scanf("%d",&p[j]);}for(int j = 0 ; j < t ; ++j){for(int r = 0 ; r < j ; ++r){graph[p[j]][p[r]] = graph[p[r]][p[j]] = 0 ;}}}int sum = prim(n) ;if(sum >= INF){puts("-1") ;}else{printf("%d\n",sum) ;}}return 0 ;}Kruskal算法代码：#include <cstdio>#include <algorithm>#include <cstring>#define MAX 550#define INF 1000000000using namespace std ;struct Edge{int x , y , w ;}edge[MAX*MAX];bool cmp(const Edge &e1 , const Edge &e2){return e1.w<e2.w ;}int f[MAX] ;int find(int x){int r = x ;while(r != f[r]){r = f[r] ;}int temp ;while(x != f[x] ){temp = f[x] ;f[x] = r ;x = temp ;}return r ;}void init(){for(int i = 0 ; i < MAX ; ++i){f[i] = i ; }}int kruskal(int n , int m){sort(edge,edge+n,cmp) ;init();int sum = 0 ,count = 0;for(int i = 0 ; i < n ; ++i){int x = find(edge[i].x) , y = find(edge[i].y) ;if(x != y){count++ ;sum += edge[i].w;f[x] = f[y] ;}}if(count<m-1){return INF ;}return sum ;}int main(){int t ;scanf("%d",&t);while(t–){int n , m , k;scanf("%d%d%d",&n,&m,&k) ;int index = 0 ;for(int i = 0 ; i < m ; ++i){int x, y ,w ;scanf("%d%d%d",&x,&y,&w) ;edge[index].x = x , edge[index].y = y , edge[index++].w = w ;}int p[MAX] ;for(int i = 0 ; i < k ; ++i){int t ;scanf("%d",&t);for(int j = 0 ; j < t ; ++j){scanf("%d",&p[j]);}for(int j = 0 ; j < t ; ++j){for(int r = 0 ; r < j ; ++r){edge[index].x = p[j] , edge[index].y = p[r] , edge[index++].w = 0 ;}}}int sum = kruskal(index,n) ;if(sum >= INF){puts("-1") ;}else{printf("%d\n",sum) ;}}return 0 ;}与君共勉
我喜欢旅游，喜欢离开自己过腻歪的城市，
