Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 13050 | Accepted: 6995 |
Description
Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.
Input
The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.
Output
Sample Input
1 0 2 3 1 2 37 2 1 17 1 2 68 3 7 1 2 19 2 3 11 3 1 7 1 3 5 2 3 89 3 1 91 1 2 32 5 7 1 2 5 2 3 7 2 4 8 4 5 11 3 5 10 1 5 6 4 2 12 0
Sample Output
0 17 16 26
#include <stdio.h>
#include <algorithm>
using namespace std;
struct edge{
int from, to, val;
bool operator < (const edge& x) const{
return val < x.val;
}
}e[11111];
int p[111];
int findset(int x){
return p[x] == x ? x : p[x] = findset(p[x]);
}
int main(){
int n, m, u, v;
while(scanf("%d", &n) != EOF){
if(n == 0) return 0;
scanf("%d", &m);
for(int i = 1; i <= m; ++i){
scanf("%d %d %d", &e[i].from, &e[i].to, &e[i].val);
}
sort(e + 1, e + 1 + m);
for(int i = 1; i <= n; ++i){
p[i] = i;
}
int ans = 0;
for(int i = 1; i <= m; ++i){
u = findset(e[i].from);
v = findset(e[i].to);
if(u != v){
p[u] = v;
ans += e[i].val;
}
}
printf("%d\n", ans);
}
}
/*
題意:
50個點,很多條可選邊,求最小生成樹。
思路:
最小生成樹裸題。並查集維護一下集合。掃一遍所有邊即可。
*/