题目:
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2566 Accepted Submission(s): 904
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
思路:
代码:
#include <iostream> #include <vector> #include <algorithm> using namespace std; const int size = 1e4 + 1; struct edge { int a, b, c; }; struct edge edges[size]; int un[size]; int cmp(edge x, edge y) { return x.c < y.c; } int find(int x) { if(x == un[x]) return x; return find(un[x]); } void Union(int x, int y) { int c1 = find(x); int c2 = find(y); un[c1] = c2; } int main() { int N, M; int S, T, C; while(cin >> N >> M) { for(int i = 0; i < N; i++) un[i] = i; for(int i = 0; i < M; i++) { cin >> edges[i].a >> edges[i].b >> edges[i].c; } sort(edges, edges+M, cmp); int count = N; int ans = 0; for(int i = 0; i < M; i++) { if(find(edges[i].a) == find(edges[i].b)) continue; Union(edges[i].a, edges[i].b); count--; ans+=edges[i].c; } if(count != 1) { printf("impossible\n\n"); continue; } printf("%d\n\n", ans); } return 0; }
原文地址:https://www.cnblogs.com/w-j-c/p/9218954.html