Ice_cream’s world III
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1146 Accepted Submission(s): 379
Problem Description
ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every city to the capital. The project’s cost should be as less as better.
Input
Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.
Output
If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.
Sample Input
2 1
0 1 10
4 0
Sample Output
10
impossible
Author
Wiskey
Source
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//并查集实现克鲁斯卡尔；
1 #include <stdio.h> 2 #include <algorithm> 3 using namespace std; 4 int father[1010] ; 5 6 struct rode 7 { 8 int a, b ,c; 9 }; 10 rode num[10010]; 11 12 bool cmp(rode a, rode b) 13 { 14 return a.c < b.c ; 15 } 16 17 int find(int a) 18 { 19 while(a != father[a]) 20 a = father[a]; 21 return a; 22 } 23 24 void mercy(int a, int b) 25 { 26 int q = find(a); 27 int p = find(b); 28 if(q != p) 29 father[q] = p; 30 31 } 32 33 int main() 34 { 35 int i, n, m; 36 while(~scanf("%d %d", &n, &m)) 37 { 38 for(i=0; i<n; i++) 39 father[i] = i; 40 for(i=0; i<m; i++) 41 scanf("%d %d %d",&num[i].a, &num[i].b, &num[i].c); 42 sort(num, num+m, cmp); 43 int total = 0; 44 for(i=0; i<m; i++) 45 { 46 //mercy(num[i].a, num[i].b); 47 if(find(num[i].a) == find(num[i].b)) 48 continue; 49 else 50 { 51 //printf("1 "); 52 total += num[i].c; 53 mercy(num[i].a, num[i].b) ; 54 } 55 } 56 int ac = 0; 57 for(i=0; i<n; i++) 58 { 59 if(father[i] == i) 60 ac++; 61 } 62 if(ac == 1) 63 printf("%d ", total); 64 else 65 printf("impossible "); 66 } 67 }