标签:-- cloc ble define i++ fine print read 优化
二分答案之后, 设答案值为v, x为小于等于v的边的数量, y为大于v的边的数量,
列出方程 x > (x + y) / k == (k - 1) * x - y > 0, 感觉有点像线性规划的样子。
然后用树形dp去check, 判合不合法的时候可以优化成一个R, 但是不优化也能过。
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned 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 ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = (int)1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;} //mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n; int L, R, K; int dp[N][51]; vector<PII> G[N]; struct Edge { int u, v, w; void read() { scanf("%d%d%d", &u, &v, &w); } } e[N]; bool dfs(int u, int fa) { for(int i = 1; i <= R; i++) { dp[u][i] = -inf; } dp[u][0] = 0; for(auto &e : G[u]) { int v = e.se, w = e.fi; if(v == fa) continue; if(dfs(v, u)) return true; for(int i = 0; i < R; i++) { for(int j = max(0, L - i - 1); i + j + 1 <= R; j++) { if(dp[u][i] + dp[v][j] + w > 0) { return true; } } } for(int i = 1; i <= R; i++) { chkmax(dp[u][i], dp[v][i - 1] + w); } } return false; } bool check(int val) { for(int i = 1; i <= n; i++) { G[i].clear(); } for(int i = 1; i < n; i++) { if(e[i].w <= val) { G[e[i].u].push_back(mk(K - 1, e[i].v)); G[e[i].v].push_back(mk(K - 1, e[i].u)); } else { G[e[i].u].push_back(mk(-1, e[i].v)); G[e[i].v].push_back(mk(-1, e[i].u)); } } return dfs(1, 0); } int main() { int T; scanf("%d", &T); while(T--) { scanf("%d", &n); for(int i = 1; i < n; i++) { e[i].read(); } scanf("%d%d%d", &K, &L, &R); int low = 1, high = 1000000000, mid, ans = -1; while(low <= high) { mid = low + high >> 1; if(check(mid)) ans = mid, high = mid - 1; else low = mid + 1; } printf("%d\n", ans); } return 0; } /* (k - 1) * x - y > 0 */
HDU - 5854 K-th value 二分 + dp (看题解)
标签:-- cloc ble define i++ fine print read 优化
原文地址:https://www.cnblogs.com/CJLHY/p/11437381.html