題目藍鏈

Description

給你一棵樹，樹上每條邊都有一個邊權。你要在上面選出\(m\)條沒有重複邊的路徑，使得選出的最短路徑儘量的長

Solution

最短的最長，這顯然就是二分答案

然後就直接在樹上貪心就可以了，對於每一個點把它的字樹儘可能多的兩兩匹配，最後如果有匹配不了的就與當前點連向父親的邊連起來，這個匹配可以用\(set\)去實現

Code

#include <bits/stdc++.h>

using namespace std;

#define fst first
#define snd second
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

inline int read() {
    int sum = 0, fg = 1; char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
    for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
    return fg * sum;
}

typedef set<pii>::iterator It;

const int maxn = 5e4 + 10;

int n, m, cnt, ans, Begin[maxn], Next[maxn << 1], To[maxn << 1], w[maxn << 1], e ;

void Add(int x, int y, int z) { To[++e] = y, Next[e] = Begin[x], Begin[x] = e, w[e] = z; }

int S[maxn], len[maxn];
bool v[maxn];
set<pii> Set;

void dfs(int now, int f) {

    for (int i = Begin[now]; i + 1; i = Next[i]) {
        int son = To[i];
        if (son == f) continue;
        dfs(son, now);
    }

    Set.clear(), S[0] = 0;
    for (int i = Begin[now]; i + 1; i = Next[i]) {
        int son = To[i];
        if (son == f) continue;
        S[++S[0]] = len[son] + w[i];
    }

    sort(S + 1, S + S[0] + 1);
    It lst = Set.begin();
    for (int i = 1; i <= S[0]; i++) {
        if (S[i] >= ans) { ++cnt, v[i] = 1; continue; }
        lst = Set.insert(lst, (pii){S[i], i}), v[i] = 0;
    }

    for (int i = 1; i <= S[0]; i++) {
        if (v[i]) continue;
        It pos = Set.lower_bound((pii){ans - S[i], 0});
        if (pos == Set.end() || (*pos) == (pii){S[i], i}) continue;
        v[i] = 1, v[pos->snd] = 1, ++cnt;
        Set.erase((pii){S[i], i}), Set.erase(pos);
    }

    len[now] = 0;
    for (int i = S[0]; i >= 1; i--)
        if (!v[i]) {
            len[now] = S[i];
            break;
        }
}

bool check(int mid) {
    ans = mid, cnt = 0;
    dfs(1, 0);
    return cnt >= m;
}

int main() {
    freopen("track.in", "r", stdin);
    freopen("track.out", "w", stdout);

    e = -1, memset(Begin, -1, sizeof Begin);
    n = read(), m = read();
    for (int i = 1; i < n; i++) {
        int x = read(), y = read(), z = read();
        Add(x, y, z), Add(y, x, z);
    }

    int l = 1, r = 5e8;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) l = mid + 1;
        else r = mid - 1;
    }

    cout << r << endl;

    return 0;
}
 

Summary

這道題一定要注意\(set\)的使用方法 考場上寫掛了

要先用一個數組把\(set\)裡的元素存起來，然後去遍歷這個陣列。在\(set\)中刪除一個元素的時候，直接對這個元素的編號打一個\(vis\)標記就可以了