洛谷 P5021 [NOIP2018]赛道重建

洛谷 P5021 [NOIP2018]赛道重建

传送门

思路

思路就是常规的思路，所以就不说了……我就是来记录一下我的\(AC\)之路的，真的是太爽了

没错……我也是一个个打的部分分，最后终于AC的，至于为什么中间又会有\(35\)、\(25\)、\(0\)这样的分数……纯粹是因为我犯了zz错误……

代码

1、\(b_i = a_i + 1\) 链的情况

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

const int N = 50011;
int a[N], n, m, cnt, head[N], sum;

struct node {
    int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
    e[++cnt].to = to;
    e[cnt].val = w;
    e[cnt].nxt = head[from];
    head[from] = cnt;
}

void dfs(int x, int fa) {
    for(int i = head[x]; i ; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fa) continue;
        dfs(y, x);
        a[x] = e[i].val;
    }
}

int check(int k) {
    int t = 0, now = 1;
    for(int i = 1; i < n; i++) {
        if(now + a[i] >= k) {
            now = 0;
            t++;
        }
        else now += a[i];
    }
    return t >= m;
}

int main() {
    n = read(), m = read();
    for(int i = 1; i < n; i++) {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        add(v, u, w);
        sum += w;
    }
    dfs(1, 0);
    int l = 1, r = sum, mid;
    while(l < r) {
        mid = (l + r) >> 1;
        if(check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l << '\n';
}

2、\(m = 1\) 求树的直径

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

const int N = 50011;
int a[N], n, m, cnt, head[N], sum, ans;

struct node {
    int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
    e[++cnt].to = to;
    e[cnt].val = w;
    e[cnt].nxt = head[from];
    head[from] = cnt;
}

int dfs(int x,int fa) {
    int sum1 = 0, sum2 = 0;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fa) continue;
        sum2 = max(sum2, dfs(y, x) + e[i].val);
        if(sum2 > sum1) swap(sum1, sum2);
    }
    ans = max(ans, sum1 + sum2);
    return sum1;
}

int main() {
    n = read(), m = read();
    for(int i = 1; i < n; i++) {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        add(v, u, w);
        sum += w;
    }
    dfs(1, 0);
    cout << ans << '\n';
}

3、\(a_i = 1\) 菊花图

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

const int N = 50011;
const int inf = 0x3f3f3f3f;
int a[N], n, m, cnt, head[N], sum, ans;

struct node {
    int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
    e[++cnt].to = to;
    e[cnt].val = w;
    e[cnt].nxt = head[from];
    head[from] = cnt;
}

bool cmp(int a, int b) {
    return a > b;
}

int main() {
    n = read(), m = read();
    for(int i = 1; i < n; i++) {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        add(v, u, w);
        sum += w;
    }
    for(int i = head[1], y; i; i = e[i].nxt) {
        y = e[i].to;
        a[y - 1] = e[i].val;
    }
    sort(a + 1, a + n, cmp);
    int ans = inf;
    for(int i = 1; i <= m; i++)
        ans = min(ans, a[i] + a[2 * m - i + 1]);
    cout << ans << '\n';
    return 0;
}

4、混起来的部分分

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

const int N = 50011;
const int inf = 0x3f3f3f3f;
int a[N], n, m, cnt, head[N], sum, ans;

struct node {
    int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
    e[++cnt].to = to;
    e[cnt].val = w;
    e[cnt].nxt = head[from];
    head[from] = cnt;
}

namespace subtask1 {
    int a[N];
    void dfs(int x, int fa) {
        for(int i = head[x]; i ; i = e[i].nxt) {
            int y = e[i].to;
            if(y == fa) continue;
            dfs(y, x);
            a[x] = e[i].val;
        }
    }

    int check(int k) {
        int t = 0, now = 1;
        for(int i = 1; i < n; i++) {
            if(now + a[i] >= k) {
                now = 0;
                t++;
            } else now += a[i];
        }
        return t >= m;
    }

    void solve() {
        dfs(1, 0);
        int l = 1, r = sum, mid;
        while(l < r) {
            mid = (l + r + 1) >> 1;
            if(check(mid)) l = mid;
            else r = mid - 1;
        }
        cout << l << '\n';
    }
}

namespace subtask2 {
    int dfs(int x,int fa) {
        int sum1 = 0, sum2 = 0;
        for(int i = head[x]; i; i = e[i].nxt) {
            int y = e[i].to;
            if(y == fa) continue;
            sum2 = max(sum2, dfs(y, x) + e[i].val);
            if(sum2 > sum1) swap(sum1, sum2);
        }
        ans = max(ans, sum1 + sum2);
        return sum1;
    }
    void solve() {
        dfs(1, 0);
        cout << ans << '\n';
    }
}

namespace subtask3 {
    bool cmp(int a, int b) {
        return a > b;
    }
    void solve() {
        for(int i = head[1], y; i; i = e[i].nxt) {
            y = e[i].to;
            a[y - 1] = e[i].val;
        }
        sort(a + 1, a + n, cmp);
        int ans = inf;
        for(int i = 1; i <= m; i++)
            ans = min(ans, a[i] + a[2 * m - i + 1]);
        cout << ans << '\n';
    }
}

int main() {
    n = read(), m = read();
    int flag = 1, f = 1;
    for(int i = 1; i < n; i++) {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        add(v, u, w);
        if(u != 1) flag = 0;
        if(v != u + 1) f = 0;
        sum += w;
    }
    if(flag) {
        subtask3::solve();
    }
    else if(f){
        subtask1::solve();
    }
    else {
        subtask2::solve();
    }
    return 0;
}

5、正解！！（\(multiset\)）

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

const int N = 50011;
const int inf = 0x3f3f3f3f;
int a[N], n, m, cnt, head[N], ans, up;

struct node {
    int to, nxt, val;
} e[N << 1];

multiset<int> s[N];
multiset<int>::iterator it;

inline void add(int from, int to, int w) {
    e[++cnt].to = to;
    e[cnt].val = w;
    e[cnt].nxt = head[from];
    head[from] = cnt;
}

int dfs(int x, int fa, int k) {
    s[x].clear();
    int w;
    for(int i =  head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fa) continue;
        w = dfs(y, x, k) + e[i].val;
        if(w >= k) ans++;
        else s[x].insert(w);
    }
    int maxn = 0;
    while(!s[x].empty()) {
        if(s[x].size() == 1) {
            return max(maxn, *s[x].begin());
        }
        it = s[x].lower_bound(k - *s[x].begin());
        if(it == s[x].begin() && s[x].count(*it) == 1) it++;
        if(it == s[x].end()) {
            maxn = max(maxn, *s[x].begin());
            s[x].erase(s[x].find(*s[x].begin()));
        } else {
            ans++;
            s[x].erase(s[x].find(*it));
            s[x].erase(s[x].find(*s[x].begin()));
        }
    }
    return maxn;
}

int dfs1(int x,int fa) {
    int sum1 = 0, sum2 = 0;
    for(int i = head[x], y; i; i = e[i].nxt) {
        y=e[i].to;
        if(y == fa) continue;
        sum2 = max(sum2, dfs1(y, x) + e[i].val);
        if(sum1 < sum2) swap(sum1, sum2);
    }
    up = max(up, sum1 + sum2);
    return sum1;
}

int check(int k) {
    ans = 0;
    dfs(1, 0, k);
    if(ans >= m) return 1;
    return 0;
}

int main() {
    n = read(), m = read();
    for(int i = 1; i < n; i++) {
        int u = read(), v = read(), w = read();
        add(u, v, w);
        add(v, u, w);
    }
    dfs1(1, 0);
    int l = 1, r = up, mid;
    while(l < r) {
        mid = (l + r + 1) >> 1;
        if(check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l << '\n';
}