【Codeforces】2015-2016 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2015) A Adjoin the Netwo

题目链接：http://codeforces.com/gym/100781/attachments

题目大意：给你n个点和l条边，让你添加一些边形成一棵树使得树的直径尽可能小。

(1 <= n <= 1e5) (0 <= l <= n - 1)

解题思路：我们先把已经形成的一些树的每个直径都求出来，把这些直径从大到小排序，如果只有1个直径，答案就是这个直径，如果有两个，就是最长和次长从中间连起来，还有一种可能就是第三长和第二长连起来比第一第二长。求树的直径，我用的是一遍dfs， dfs 的时候维护当前节点为根的最长和次长值，最后根节点最长+次长就是直径。

//2017-08-01 22:13
//2017-08-01 22:29
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cstdlib>
#include<vector>
using namespace std;
const int MaxN = 1e5;int n, m, len;
vector<int>edge[MaxN + 5];
int q[MaxN + 5];
bool vis[MaxN + 5];bool cmp(int a, int b){ return a > b; }int dfs(int u){vis[u] = true;int m1 = 0, m2 = 0;for(int i = 0; i < edge[u].size(); i++){int v = edge[u][i];if(!vis[v]){int tmp = dfs(v) + 1;if(tmp > m1){m2 = m1, m1= tmp;}else if(tmp > m2){m2 = tmp;}}}len = max(len, m1 + m2);return m1;
}int main(){scanf("%d %d", &n, &m);for(int i = 1; i <= m; i++){int u, v;scanf("%d %d", &u, &v);edge[u].push_back(v);edge[v].push_back(u);}int cnt = 0, ans = 0;for(int i = 0; i < n; i++){len = 0;if(!vis[i]){int tmp = dfs(i);q[++cnt] = len;ans = max(ans, len);}}sort(q + 1, q + 1 + cnt, cmp);if(cnt > 1) ans = max(ans, (q[1] + 1) / 2 + (q[2] + 1) / 2 + 1);if(cnt > 2) ans = max(ans, (q[2] + 1) / 2 + (q[3] + 1) / 2 + 2);printf("%d\n", ans);return 0;
}

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【Codeforces】2015-2016 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2015) A Adjoin the Netwo

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