杭电2988--Dark roads

Dark roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 798 Accepted Submission(s): 329

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

Source

2009/2010 Ulm Local Contest

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RE: 开始时候想多了，排了两次序，修路就是修路，只不过让你找花费最少的。 map[][] . , .

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 using namespace std;
 7 int father[200200];
 8 int n, m;
 9 struct rode
10 {
11     int x, y, w;
12 } num[200200], shu[200200];
13 
14 bool cmp(rode x, rode y)
15 {
16     return x.w > y.w;
17 }
18 
19 bool cpm(rode x, rode y)
20 {
21     return x.w < y.w;
22 }
23 
24 void init()
25 {
26     for(int i = 0; i < n; i++)
27         father[i] = i; 
28 }
29 
30 int find(int a)
31 {
32     if(a == father[a])
33         return a;
34     else
35         return father[a] = find(father[a]);
36 }
37 
38 bool mercy(int a, int b)
39 {
40     int q = find(a);
41     int p = find(b);
42     if(q != p)
43     {
44         father[q] = p;
45         return true;
46     }
47     else
48         return false;
49 }
50 
51 int main()
52 {
53     while(~scanf("%d %d", &n, &m), n+m)
54     {
55         init(); int sum = 0;
56         for(int i = 0; i < m; i++)
57         {
58             scanf("%d %d %d", &num[i].x, &num[i].y, &num[i].w);
59             shu[i].x = num[i].x; shu[i].y = num[i].y; shu[i].w = num[i].w;
60             sum += num[i].w;
61         }
62         /*int sum = 0;
63         sort(num, num + m, cmp);
64         for(int i = 0; i < m; i++)
65         {
66             if(mercy(num[i].x, num[i].y))
67                 sum += num[i].w;
68         }*/
69         int total = 0;
70         sort(shu, shu + m, cpm);
71         for(int i = 0; i < m; i++)
72         {
73             if(mercy(shu[i].x, shu[i].y))
74                 total += shu[i].w;
75         } 
76         printf("%d
", abs(sum - total));
77     }
78     return 0;
79 }