提交记录 47673


用户 题目 状态 得分 用时 内存 语言 代码长度
jiegec noip18c. 【NOIP2018】赛道修建 Wrong Answer 80 41.118 ms 6080 KB C++17 2.06 KB
提交时间 评测时间
2026-09-13 01:12:18 2026-09-13 01:12:24
// This code is AI-generated. (AI 生成的代码)
// NOIP2018 赛道修建: binary search the answer X.  Check(X): DFS; child chains
// reaching a node are either closed (length >= X) or paired with each other to
// reach X (greedy two-pointer); the largest unpaired chain is passed to parent.
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
enum { MAXN = 50005, MAXE = 100005 };
static int head[MAXN], nx_[MAXE], to_[MAXE], wt_[MAXE], ec;
static int X, formed;
static int pool[MAXN], sp;
static void ae(int u, int v, int w) {
    to_[++ec] = v; wt_[ec] = w; nx_[ec] = head[u]; head[u] = ec;
}
static int dfs(int u, int fa) {
    int base = sp, k = 0;
    for (int e = head[u]; e; e = nx_[e]) {
        int v = to_[e];
        if (v == fa) continue;
        int d = dfs(v, u) + wt_[e];
        if (d >= X) formed++;
        else { pool[base + k] = d; k++; sp = base + k; }
    }
    sort(pool + base, pool + base + k);
    int l = 0, r = k - 1;
    while (l < r) {
        if (pool[base + l] + pool[base + r] >= X) { formed++; l++; r--; }
        else l++;
    }
    int ret = (l == r) ? pool[base + l] : 0;
    sp = base;
    return ret;
}
static char ib[1 << 22], ob[32];
static char *gp;
static inline int rd() { while (*gp < '0') gp++; int v = 0; while (*gp >= '0' && *gp <= '9') v = v * 10 + (*gp++ - '0'); return v; }
int main() {
    int len = (int)fread(ib, 1, sizeof(ib) - 1, stdin);
    ib[len] = 0; gp = ib;
    int n = rd(), m = rd();
    long long total = 0;
    for (int i = 1; i < n; i++) {
        int a = rd(), b = rd(), w = rd();
        ae(a, b, w); ae(b, a, w); total += w;
    }
    int lo = 0, hi = (int)(total / m), ans = 0;
    while (lo <= hi) {
        int mid = (lo + hi) >> 1;
        X = mid; formed = 0;
        if (mid == 0) { ans = 0; lo = 1; continue; }
        dfs(1, 0);
        if (formed >= m) { ans = mid; lo = mid + 1; }
        else hi = mid - 1;
    }
    char *op = ob; int k = 0; char t[16];
    while (ans) { t[k++] = (char)('0' + ans % 10); ans /= 10; }
    while (k) *op++ = t[--k];
    *op++ = '\n';
    fwrite(ob, 1, op - ob, stdout);
    return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Testcase #110.66 us40 KBAcceptedScore: 5

Testcase #211.33 us40 KBAcceptedScore: 5

Testcase #310.31 us40 KBAcceptedScore: 5

Testcase #4467.54 us80 KBAcceptedScore: 5

Testcase #541.118 ms1 MB + 304 KBAcceptedScore: 5

Testcase #636.956 ms1 MB + 580 KBAcceptedScore: 5

Testcase #722.372 ms1 MB + 312 KBAcceptedScore: 5

Testcase #833.89 ms2 MB + 148 KBAcceptedScore: 5

Testcase #9255.12 us148 KBAcceptedScore: 5

Testcase #1013.913 ms3 MB + 592 KBAcceptedScore: 5

Testcase #1125.896 ms5 MB + 960 KBAcceptedScore: 5

Testcase #1221.02 us40 KBAcceptedScore: 5

Testcase #1319.06 us40 KBAcceptedScore: 5

Testcase #1456.73 us44 KBAcceptedScore: 5

Testcase #1554.84 us48 KBAcceptedScore: 5

Testcase #16292.85 us88 KBAcceptedScore: 5

Testcase #17331.51 us76 KBWrong AnswerScore: 0

Testcase #1821.416 ms1 MB + 312 KBWrong AnswerScore: 0

Testcase #1918.608 ms1 MB + 572 KBWrong AnswerScore: 0

Testcase #2035.996 ms2 MB + 640 KBWrong AnswerScore: 0


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