首页 > ACM题库 > HDU-杭电 > HDU 2988-Dark roads-最小生成树-[解题报告]HOJ
2014
02-24

HDU 2988-Dark roads-最小生成树-[解题报告]HOJ

Dark roads

问题描述 :

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?

输入:

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.

输出:

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.

样例输入:

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

样例输出:

51

http://acm.hdu.edu.cn/showproblem.php?pid=2988

最小生成树啦。赤裸裸的Kruskal算法。这个题不能用Prim,因为没办法把二维数组开那么大哦。

嘿嘿~~ 还是比较简单的。

#include <iostream>
#include <algorithm>
using namespace std;
#define N 200005
int junctions[N];

struct road_node{
	int x,y,len;
};
road_node roads[N];

int findfather(int n){
	if (n!=junctions[n])
		junctions[n]=findfather(junctions[n]);
	return junctions[n];
}

bool cmp(road_node a,road_node b){
	return a.len<b.len;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("2988in.txt","r",stdin);
#endif
	int n,m,i,a,b,len_total,num,light;
	while (scanf("%d%d",&n,&m)!=EOF){
		if (n==0&&m==0)
			break;
		len_total=0;
		num=0;
		for (i=0;i<n;i++)
			junctions[i]=i;
		for (i=0;i<m;i++){
			scanf("%d%d%d",&roads[i].x,&roads[i].y,&roads[i].len);
			len_total+=roads[i].len;
		}
		sort(roads,roads+m,cmp);
		i=0;
		light=0;
		while (num<n-1&&i<m){
			a=findfather(roads[i].x);
			b=findfather(roads[i].y);
			if (a!=b){
				junctions[b]=a;
				num++;
				light+=roads[i].len;
			}
			i++;
		}
		printf("%d\n",len_total-light);
	}
	return 0;
}

解题参考:http://blog.csdn.net/operator456/article/details/8560895


  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  2. 题本身没错,但是HDOJ放题目的时候,前面有个题目解释了什么是XXX定律。
    这里直接放了这个题目,肯定没几个人明白是干啥

  3. #include <stdio.h>
    int main(void)
    {
    int arr[] = {10,20,30,40,50,60};
    int *p=arr;
    printf("%d,%d,",*p++,*++p);
    printf("%d,%d,%d",*p,*p++,*++p);
    return 0;
    }

    为什么是 20,20,50,40,50. 我觉得的应该是 20,20,40,40,50 . 谁能解释下?