http://codeforces.com/problemset/problem/348/B
注意到如果顶点的数值确定了，那么它分下去的个数也就确定了，那么可以暴力枚举顶点的数值。
顶点的数值是和LCM相隔的，LCM就是，比如1有三个子节点，那么1的数值起码都是3的倍数，不然不能整除。
同理，1有三个儿子2、3、4、，如果3有三个儿子，那么1就要起码是9的倍数了，因为需要分给3的时候至少是3.
所以算出整颗树的LCM，叶子节点LCM是1，其他的LCM = lcm(所有子节点) * son[cur]
算出这个后，就可以暴力枚举顶点1的值了，每次枚举都dfs一次，有点暴力1700ms，学学正解先。
注意一点的就是LCM会爆LL，这个时候直接输出sum即可
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> const int maxn = 1e5 + 20; int a[maxn]; LL sum; struct Node { int u, v, tonext; }e[maxn * 2]; int first[maxn], num; void addEdge(int u, int v) { ++num; e[num].u = u, e[num].v = v, e[num].tonext = first[u]; first[u] = num; } int vis[maxn], DFN = 1; int son[maxn]; void findSon(int cur) { for (int i = first[cur]; i; i = e[i].tonext) { int v = e[i].v; if (vis[v] == DFN) continue; vis[v] = DFN; son[cur]++; findSon(v); } } LL ans = 1e18L, tans; LL LCM; LL lcm(LL a, LL b) { return a / __gcd(a, b) * b; } bool flag; void dfs(int cur, LL val) { if (flag) return; if (son[cur] == 0) { if (val > a[cur]) { flag = true; return; } tans += a[cur] - val; return; } for (int i = first[cur]; i; i = e[i].tonext) { int v = e[i].v; if (vis[v] == DFN) continue; vis[v] = DFN; if (val % son[cur] != 0) { flag = true; return; } dfs(v, val / son[cur]); } } LL calc(int cur) { if (son[cur] == 0) return 1; LL thisLCM = 1; for (int i = first[cur]; i; i = e[i].tonext) { int v = e[i].v; if (vis[v] == DFN) continue; vis[v] = DFN; thisLCM = lcm(thisLCM, calc(v)); } if (thisLCM > sum / son[cur]) { printf("%I64d ", sum); exit(0); } return son[cur] * thisLCM; } void work() { int n; scanf("%d", &n); LCM = 1; for (int i = 1; i <= n; ++i) { scanf("%d", &a[i]); sum += a[i]; } for (int i = 1; i <= n - 1; ++i) { int u, v; scanf("%d%d", &u, &v); addEdge(u, v); addEdge(v, u); } vis[1] = DFN; findSon(1); ++DFN; vis[1] = DFN; LCM = calc(1); // cout << LCM << endl; ans = sum; sum /= LCM; sum *= LCM; for (LL i = sum; i >= LCM; i -= LCM) { ++DFN, tans = 0; flag = false; // dfs(1, 24); vis[1] = DFN; dfs(1, i); if (flag) continue; printf("%I64d ", ans - i); return; } printf("%I64d ", ans); } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif work(); return 0; }
因为已经知道根节点的值是LCM、2 * LCM、3 * LCM、..... sum中合法的最大的哪一个，那么可以二分答案。
一开始想把她们全部存入vector再二分，但是发现不用，直接二分id就行，就是二分LCM上面的那个倍数就可以了。
500ms
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> const int maxn = 1e5 + 20; int a[maxn]; LL sum; struct Node { int u, v, tonext; }e[maxn * 2]; int first[maxn], num; void addEdge(int u, int v) { ++num; e[num].u = u, e[num].v = v, e[num].tonext = first[u]; first[u] = num; } int vis[maxn], DFN = 1; int son[maxn]; void findSon(int cur) { for (int i = first[cur]; i; i = e[i].tonext) { int v = e[i].v; if (vis[v] == DFN) continue; vis[v] = DFN; son[cur]++; findSon(v); } } LL ans = 1e18L, tans; LL LCM; LL lcm(LL a, LL b) { return a / __gcd(a, b) * b; } bool flag; void dfs(int cur, LL val) { if (flag) return; if (son[cur] == 0) { if (val > a[cur]) { flag = true; return; } tans += a[cur] - val; return; } for (int i = first[cur]; i; i = e[i].tonext) { int v = e[i].v; if (vis[v] == DFN) continue; vis[v] = DFN; if (val % son[cur] != 0) { flag = true; return; } dfs(v, val / son[cur]); } } LL calc(int cur) { if (son[cur] == 0) return 1; LL thisLCM = 1; for (int i = first[cur]; i; i = e[i].tonext) { int v = e[i].v; if (vis[v] == DFN) continue; vis[v] = DFN; thisLCM = lcm(thisLCM, calc(v)); } if (thisLCM > sum / son[cur]) { printf("%I64d ", sum); exit(0); } return son[cur] * thisLCM; } void work() { int n; scanf("%d", &n); LCM = 1; for (int i = 1; i <= n; ++i) { scanf("%d", &a[i]); sum += a[i]; } for (int i = 1; i <= n - 1; ++i) { int u, v; scanf("%d%d", &u, &v); addEdge(u, v); addEdge(v, u); } vis[1] = DFN; findSon(1); ++DFN; vis[1] = DFN; LCM = calc(1); // cout << LCM << endl; ans = sum; sum /= LCM; sum *= LCM; LL be = 0, en = sum / LCM; while (be <= en) { LL mid = ((be + en) >> 1); ++DFN, flag = false; vis[1] = DFN; dfs(1, mid * LCM); if (flag) { en = mid - 1; } else be = mid + 1; } cout << ans - LCM * en << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif work(); return 0; }