NCPC 2015 - Problem A - Adjoin the Networks
题目链接 : http://codeforces.com/gym/100781/attachments
题意 :
有n个编号为0-n-1的点, 给出目前已经有的边(最多n-1条), 问如何添加最少的边, 使得整个图连通, 且其中两点间距离的最大值最小, 一条边距离为1单位
思路 :
两点间距离的情况 : 1. 子图中任意两点间距离 2.两个子图中两点间的距离
无论如何添加边对第一种的距离都没有影响, 对第二种却有影响
考虑添加一条边连通两个子图后, 最长的距离为 子图1中距离添加边的节点的最长的距离 + 子图2中距离添加边的节点的最长的距离 + 添加的1条边的距离
所以选择添加边的节点是 在某个子图中, 使所有节点到它的最长距离最小的点
但是不用算出这个点, 只需要这个距离, 这个距离就是一个子图中距离最长两点距离的一半, 即(max_length + 1) / 2
一个图中最长距离求法是在对某个子图任一节点深搜, 搜到距离最长的点, 再对这个点深搜, 记录最长距离(新技能)
最终取 : 情况1 和 情况2 距离最大值
注意情况二中, 如果存在三个子图最大(max_length + 1) / 2 都相等, 最终答案是要加2条边的距离而不是加1条, 比如0-1 1-2 3-4 这三个子图
- #include <cstdio>
- #include <cstring>
- #include <algorithm>
- #include <vector>
- using namespace std;
- const int MAXN = 1e5+;
- vector<int> edge[MAXN];
- int depth[MAXN];
- bool vis[MAXN];
- int maxx, maxx1, mark1;
- int length[MAXN];
- int Max(int a, int b)
- {
- return a > b ? a : b;
- }
- void Dfs(int u, int fa)
- {
- vis[u] = ;
- depth[u] = depth[fa] + ;
- if(depth[u] > maxx1) {
- mark1 = u;
- maxx1 = depth[u];
- }
- int len = edge[u].size();
- for(int i = ; i < len; i++) {
- int v = edge[u][i];
- if(v == fa) continue;
- Dfs(v, u);
- }
- }
- bool cmp(int a, int b)
- {
- return a > b;
- }
- void Init(int n)
- {
- for(int i = ; i <= n; i++) {
- edge[i].clear();
- vis[i] = ;
- }
- maxx = ;
- }
- int main()
- {
- int n, l;
- int u, v;
- while(scanf("%d %d", &n, &l) != EOF) {
- Init(n);
- for(int i = ; i < l; i++) {
- scanf("%d %d", &u, &v);
- edge[u].push_back(v);
- edge[v].push_back(u);
- }
- int cnt = ;
- for(int i = ; i < n; i++) {
- maxx1 = ;
- if(vis[i] == ) {
- depth[i] = -;
- Dfs(i, i);
- depth[mark1] = -;
- if(maxx1 != ) {
- Dfs(mark1, mark1);
- }
- length[i] = (maxx1 + ) / ;
- if(maxx1 > maxx) maxx = maxx1;
- cnt++;
- }
- }
- sort(length, length+n, cmp);
- if(n == ) printf("0\n");
- else if(n == ) printf("1\n");
- else if(n == ) printf("2\n");
- else if(cnt == ) printf("%d\n", maxx);
- else if(length[] == length[] && length[] == length[]) {
- printf("%d\n", Max(maxx, length[] + length[] + ));
- }
- else {
- printf("%d\n", Max(maxx, length[] + length[] + ));
- }
- }
- return ;
- }
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