POJ 1287:Networking(最小生成树Kruskal)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 5976 | Accepted: 3231 |
Description
given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the
area.
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 following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only
one number P=0 denotes the end of the input. The data sets are separated with an empty line.
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
题意:
P个点。R对边。找出最小通路和
分析:
长度为边权的最小生成树问题
kruskal & 并查集
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f using namespace std; const int maxn=5100; int r[maxn];
int n, m; struct node
{
int x;
int y;
int cost;
};
node a[maxn]; bool cmp ( node a, node b )
{
return a.cost<b.cost;
} void init()
{
for( int i=0; i<n; i++ )
scanf( "%d%d%d", &a[i].x, &a[i].y, &a[i].cost );
sort( a, a+n, cmp );
for( int i=0; i<n; i++ )
r[i] = i;
} int find( int x )
{
return r[x]==x? x : r[x] = find( r[x] );
} void kruskal()
{
int cnt = 0;
int ans = 0;
int x; int y;
for( int i=0; i<n; i++ )
{
x = find( a[i].x );
y = find( a[i].y );
if( x!=y )
{
cnt++;
ans += a[i].cost;
if( cnt == m-1 )
break;
r[y] = x;
}
}
printf( "%d\n", ans );
} int main()
{
while( scanf( "%d", &m )==1 &&m )
{
scanf( "%d", &n );
init();
kruskal();
}
return 0;
}
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