3427: Dark roads

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本文探讨了在不牺牲市民安全感的前提下，通过优化道路照明来减少能耗的方法，具体介绍了如何在夜晚关闭某些街道的照明，同时确保从每个交汇点到其他交汇点至少有一条被照亮的道路，以此达到节省开支的目的。

http://cs.scu.edu.cn/soj/problem.action?id=3427

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 Specification

The input file contains several test cases. Each test case starts with two numbersm andn, the number of junctions in Byteland and the number of roads in Byteland, respectively. Input is terminated bym=n=0. Otherwise,1 ≤ m ≤ 200000 andm-1 ≤ n ≤ 200000. Then follown integer triplesx, y, z specifying that there will be a bidirectional road betweenx andy with lengthz meters (0 ≤ x, y < m andx ≠ 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 Specification

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

Sample Input

Sample Output
51
￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥做了这道题以后感觉自己对kruskal算法又有了一个新的认识，以前的充其量只能
算是模板，真正自己敲得就是这道题了！！嘎嘎！！奖励自己一下！
#include<stdio.h>
#include<iostream>
#include<algorithm>
#define NN 200000000
using namespace std;
struct node
{
int l,r,value;
}g[200005];
int cmp(const node&a,const node&b)
{
return a.value<b.value;
}
int set[200005];
int find(int x)
{
if(x!=set[x])
set[x]=find(set[x]);
return set[x];
}
void merge(int x,int y)
{
int fx,fy;
fx=find(x);
fy=find(y);
if(fx==fy)
return ;
if(fx>fy)
set[fx]=fy;
else
set[fy]=fx;
}
int main()
{
int i,j,k,m,n,x,y,z,ww;
int map[105][105];
while(scanf("%d%d",&m,&n),m,n)
{
for(i=0;i<m;i++)
   set[i]=i;
   ww=0;
for(i=0;i<n;i++)
{
   scanf("%d%d%d",&g[i].l,&g[i].r,&g[i].value);
   ww+=g[i].value;
   }
sort(g,g+n,cmp);
/printf("\n");
for(i=0;i<n;i++)
printf("%d %d %d\n",g[i].l,g[i].r,g[i].value);
   printf("\n");/
int sum=0;
for(i=0;i<n;i++)
{
if(find(g[i].l)!=find(g[i].r))
{
merge(g[i].l,g[i].r);
sum+=g[i].value;
//printf("%d %d %d\n",g[i].l,g[i].r,g[i].value);
}
}
printf("%d\n",ww-sum);
}
return 0;
}