kruskal算法

Dark roads

Time Limit: 2000ms, Special Time Limit: 5000ms, Memory Limit: 65536KB

Total submit users: 15, Accepted users: 15

Problem 10648 : No special judgement

Problem description

  Economic times these days are tough, even in Byteland. To reduce the operating costs, the government of Byteland has decided to optimize the road lighting. Till now every road was illuminated all night long, which costs 1 Bytelandian Dollar per meter and day. To save money, they decided to no longer illuminate every road, but to switch off the road lighting of some streets. To make sure that the inhabitants of Byteland still feel safe, they want to optimize the lighting in such a way, that after darkening some streets at night, there will still be at least one illuminated path from every junction in Byteland to every other junction.

What is the maximum daily amount of money the government of Byteland can save, without making their inhabitants feel unsafe?

Input

  The input file contains several test cases. Each test case starts with two numbers m and n , the number of junctions in Byteland and the number of roads in Byteland, respectively. Input is terminated by m=n=0 . Otherwise, 1 ≤ m ≤ 200000 and m-1 ≤ n ≤ 200000 . Then follow n integer triples x, y, z specifying that there will be a bidirectional road between x and y with length z meters (0 ≤ x, y < m and x ≠ y ). The graph specified by each test case is connected. The total length of all roads in each test case is less than 2³¹ .

Output

  For each test case print one line containing the maximum daily amount the government can save.

Sample Input

7 11

0 1 7

0 3 5

1 2 8

1 3 9

1 4 7

2 4 5

3 4 15

3 5 6

4 5 8

4 6 9

5 6 11

0 0

Sample Output

51

Problem Source

  University of Ulm Local Contest 2009

kruskal算法的典型題目；節約能源，關閉一些路的路燈，但同時要保證每個居民去是有有路燈的路相連的，求最小的方案。構建最小生成樹；

通過並查集來保證不構成環路，通過sort排序進行排序。

這簡直是一道模板性質的題目……

#include <iostream> using namespace std; int p[200000]; int rank[200000]; struct node { int s; int e; int cost; }road[200000]; int find_set(int x) { if(x!=p[x]) p[x]=find_set(p[x]); return p[x]; } void make_set(int x) { p[x]=x; rank[x]=0; } void Union(int x,int y) { x=find_set(x); y=find_set(y); if(rank[x]>rank[y]) { p[y]=x; } else { p[x]=y; if(rank[x]==rank[y]) { rank[y]++; } } } bool cmp(node a,node b) { return a.cost<b.cost; } int main() { int m,n; int i,j; unsigned int sum; int save; while(scanf("%d%d",&m,&n)) { if(m==0&&n==0) break; sum=0; save=0; for(i=0;i<m;i++) { make_set(i); } for(i=0;i<n;i++) { scanf("%d%d%d",&road[i].s,&road[i].e,&road[i].cost); sum+=road[i].cost; } sort(road,road+n,cmp); for(i=0;i<n&&m;i++) { if(find_set(road[i].s)!=find_set(road[i].e)) { Union(road[i].s,road[i].e); m--; save+=road[i].cost; } } printf("%d/n",sum-save); } return 0; }