Codeforces 1092E Minimal Diameter Forest
作者:互联网
首先我们找出每个连通块中的特殊点, 特殊点的定义是到各种个连通块中距离的最大值最小的点,
每个连通块肯定通过特殊点连到其他连通块, 我们把有最大值的特殊点当作根, 然后其他点直接接在这个点中, 形成菊花图。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 1000 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); int n, m, cnt, mx, DIA, belong[N], mn[N], who[N], id[N], dia[N]; vector<int> G[N]; void dfs(int u) { belong[u] = cnt; for(auto& v : G[u]) if(!belong[v]) dfs(v); } void dfs2(int u, int fa, int depth) { mx = max(mx, depth); for(auto& v : G[u]) if(v != fa) dfs2(v, u, depth + 1); } bool cmp(const int& x, const int& y) { return mn[x] > mn[y]; } int main() { memset(mn, inf, sizeof(mn)); scanf("%d%d", &n, &m); for(int i = 1; i <= m; i++) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } for(int i = 1; i <= n; i++) { if(!belong[i]) ++cnt, dfs(i); } for(int i = 1; i <= n; i++) { mx = 0; dfs2(i, 0, 0); if(mx < mn[belong[i]]) { mn[belong[i]] = mx; who[belong[i]] = i; } dia[belong[i]] = max(dia[belong[i]], mx); } for(int i = 1; i <= cnt; i++) { id[i] = i; DIA = max(DIA, dia[i]); } sort(id + 1, id + 1 + cnt, cmp); if(cnt >= 2) DIA = max(DIA, mn[id[1]] + mn[id[2]] + 1); if(cnt >= 3) DIA = max(DIA, mn[id[2]] + mn[id[3]] + 2); printf("%d\n", DIA); for(int i = 2; i <= cnt; i++) printf("%d %d\n", who[id[1]], who[id[i]]); return 0; } /* */
标签:Diameter,const,int,mn,Codeforces,DIA,id,Minimal,define 来源: https://www.cnblogs.com/CJLHY/p/10492626.html