AGC018D - Tree and Hamilton Path
题意
给出一个n个点的带边权的树，再给出一个n个点的完全图，其中每两个点之间的距离为这两个点在树上的距离，求最大的哈密顿图。
做法
直接考虑在树上的游历，如果存在一条边把树分成大小相同的两半，然后在两半中的点中交替走，这样子显然是最优的，因为每条边都会达到可能的最多的访问次数；否则必然存在一个点（重心），去除这个点之后的森林里每棵树的大小都不大于n/2，这样最优的游历一定是从每棵树出来走到另一棵中没走过的点，这样的安排总是存在的。考虑到最后并不会再访问重心，取重心连出去的权值最小的一条边来承担这个代价。
这种题就考虑到最优情况仔细分析就好了.
1 #include <bits/stdc++.h> 2 using namespace std; 3 namespace my_header { 4 #define pb push_back 5 #define mp make_pair 6 #define pir pair<int, int> 7 #define vec vector<int> 8 #define pc putchar 9 #define clr(t) memset(t, 0, sizeof t) 10 #define pse(t, v) memset(t, v, sizeof t) 11 #define bl puts("") 12 #define wn(x) wr(x), bl 13 #define ws(x) wr(x), pc(' ') 14 const int INF = 0x3f3f3f3f; 15 typedef long long LL; 16 typedef double DB; 17 inline char gchar() { 18 char ret = getchar(); 19 for(; (ret == '\n' || ret == '\r' || ret == ' ') && ret != EOF; ret = getchar()); 20 return ret; } 21 template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) { 22 for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar()); 23 if (c == '-') { flg = -1; c = getchar(); } 24 for(ret = 0; '0' <= c && c <= '9'; c = getchar()) 25 ret = ret * 10 + c - '0'; 26 ret = ret * flg; } 27 inline int fr() { int t; fr(t); return t; } 28 template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); } 29 template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); } 30 template<class T> inline char wr(T a, int b = 10, bool p = 1) { 31 return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 32 (wr(a/b, b, 0), pc('0' + a % b))); 33 } 34 template<class T> inline void wt(T a) { wn(a); } 35 template<class T> inline void wt(T a, T b) { ws(a), wn(b); } 36 template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); } 37 template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); } 38 template<class T> inline T gcd(T a, T b) { 39 return b == 0 ? a : gcd(b, a % b); } 40 template<class T> inline T fpw(T b, T i, T _m, T r = 1) { 41 for(; i; i >>= 1, b = b * b % _m) 42 if(i & 1) r = r * b % _m; 43 return r; } 44 }; 45 using namespace my_header; 46 47 const int N = 1e5 + 100; 48 const int M = N<<1; 49 50 struct Edge { 51 int e, h[N], t[M], n[M], w[M]; 52 void init() { 53 e = 0; memset(h, -1, sizeof h); 54 } 55 void add(int u, int v, int c) { 56 t[e] = v, n[e] = h[u], w[e] = c, h[u] = e++; 57 t[e] = u, n[e] = h[v], w[e] = c, h[v] = e++; 58 } 59 } e; 60 61 int n; 62 LL ans; 63 64 int siz[N], ef[N]; 65 void dfs(int u, int f = -1) { 66 siz[u] = 1; 67 for (int i = e.h[u]; i != -1; i = e.n[i]) { 68 int v = e.t[i]; 69 if (v == f) continue; 70 ef[v] = i; 71 dfs(v, u); 72 siz[u] += siz[v]; 73 } 74 ans += 1LL * e.w[ef[u]] * min(siz[u], n - siz[u]) * 2; 75 } 76 77 78 int main() { 79 #ifdef lol 80 freopen("d.in", "r", stdin); 81 freopen("d.out", "w", stdout); 82 #endif 83 84 fr(n); 85 e.init(); 86 for (int i = 1; i < n; ++i) { 87 int u, v, w; 88 fr(u, v, w); 89 e.add(u, v, w); 90 } 91 dfs(1); 92 //for (int i = 1; i <= n; ++i) 93 // ws(siz[i]); 94 //puts(""); 95 for (int i = 2; i <= n; ++i) { 96 if (siz[i] * 2 == n) { 97 wt(ans - e.w[ef[i]]); 98 return 0; 99 } 100 } 101 int val = INT_MAX; 102 103 //wt(ans); 104 for (int i = 1; i <= n; ++i) { 105 int maxSiz = n - siz[i], minEdge = i == 1 ? INT_MAX : e.w[ef[i]]; 106 for (int j = e.h[i]; j != -1; j = e.n[j]) { 107 if (e.t[j] != e.t[ef[i] ^ 1]) { 108 maxSiz = max(maxSiz, siz[e.t[j]]); 109 minEdge = min(minEdge, e.w[j]); 110 } 111 } 112 //wt(maxSiz); 113 if (maxSiz * 2 <= n) { 114 val = min(val, minEdge); 115 } 116 } 117 wt(ans - val); 118 119 return 0; 120 }
