Highway

Highway
Accepted : 78		Submit : 275
Time Limit : 4000 MS		Memory Limit : 65536 KB

 

In ICPCCamp there were i . It is guaranteed that any two cities reach each other using only roads. Bobo would like to build y using roads. As Bobo is rich, he would like to find the most expensive way to build the ) highways. Input The input contains zero or more test cases and is terminated by end-of-file. For each test case: The first line contains an integer i . 5 n 8 The number of test cases does not exceed 10 . Output For each test case, output an integer which denotes the result. Sample Input 5 1 2 2 1 3 1 2 4 2 3 5 1 5 1 2 2 1 4 1 3 4 1 4 5 2 Sample Output 19 15

 

 

//题意:有一棵树，问所有点到这棵树中所有点最远距离和为多少

//因为这个是树，所以，从任意一点 dfs 搜最远点，再从最远点 dfs 搜最远点，再dfs一下，这样，保存了2个最远点到任意点的距离，遍历一遍选最大的那个即可，最后减去直径，因为重复算了一次

一共就 4 n 次遍历吧，1800 ms 还行吧

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 using namespace std;
 8 #define LL long long
 9 #define INF 0x3f3f3f3f3f3f3f3f
10 #define MX 100005
11 struct To
12 {
13     int to;
14     LL c;
15 };
16 
17 int n;
18 int far;
19 LL step;
20 vector<To> road[MX];
21 LL dis[2][MX];
22 
23 void Init()
24 {
25     for (int i=1;i<=n;i++)
26         road[i].clear();
27 }
28 
29 void dfs(int x,int pre,LL s,int kk)//所在位置，从前，距离，第几次
30 {
31     if (kk!=0) dis[kk-1][x]=s;
32 
33     if (s>step)
34     {
35         far=x;
36         step=s;
37     }
38     for (int i=0;i<(int)road[x].size();i++)
39     {
40         if (road[x][i].to!=pre)
41             dfs(road[x][i].to,x,road[x][i].c+s,kk);
42     }
43 }
44 
45 int main()
46 {
47     while (~scanf("%d",&n))
48     {
49         Init();
50         for (int i=0;i<n-1;i++)
51         {
52             int a,b,c;
53             scanf("%d%d%d",&a,&b,&c);
54             road[a].push_back((To){b,c});
55             road[b].push_back((To){a,c});
56         }
57 
58         step=0;
59         dfs(1,0,0,0);
60 
61         step=0;
62         dfs(far,0,0,1);
63         step=0;
64         dfs(far,0,0,2);
65 
66         LL ans = 0;
67         for (int i=1;i<=n;i++)
68             ans += max(dis[0][i],dis[1][i]);
69         ans -= step;
70         printf("%I64d\n",ans);
71     }
72     return 0;
73 }

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