LOJ #2473. 「九省联考 2018」秘密袭击

#2473. 「九省联考 2018」秘密袭击

链接

分析：

　　首先枚举一个权值W，计算这个多少个连通块中，第k大的数是这个权值。

　　$f[i][j]$表示到第i个节点，有j个大于W数的连通块的个数。然后背包转移。

　　复杂度是$O(n^2k)$，时限5s，然后卡卡常就过了。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef unsigned int ui;
inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 2005, mod = 64123;
struct Edge{ int to, nxt; } e[N << 1];
int head[N], id[N], fa[N], st[N], ed[N], a[N];
ui f[N][N];
int n, k, w, Now, En;

inline bool cmp(int x,int y) { return a[x] == a[y] ? x <= y : a[x] < a[y]; }
inline void add_edge(int u,int v) {
    ++En; e[En].to = v, e[En].nxt = head[u]; head[u] = En;
    ++En; e[En].to = u, e[En].nxt = head[v]; head[v] = En;
}
void dfs(int u) {
    for (int i = st[u]; i <= ed[u]; ++i) f[u][i] = 0;
    if (cmp(Now, u)) st[u] = ed[u] = 1;
    else st[u] = ed[u] = 0;
    f[u][st[u]] = 1;
    for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v == fa[u]) continue;
        fa[v] = u;
        dfs(v);
        for (int j = ed[u]; j >= st[u]; --j) 
            for (int k = ed[v]; k >= st[v]; --k) 
                f[u][j + k] = (f[u][j + k] + f[u][j] * f[v][k]) % mod;
        ed[u] += ed[v];
    }
    for (int i = st[u]; i <= ed[u]; ++i) f[0][i] = (f[0][i] + f[u][i]) % mod;
}
int main() {
    n = read(), k = read(), w = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i < n; ++i) {
        int u = read(), v = read();
        add_edge(u, v);
    }
    for (int i = 1; i <= n; ++i) id[i] = i;
    sort(id + 1, id + n + 1, cmp);
    ui ans = 0;
    for (int i = 1; i <= n; ++i) {
        memset(f[0], 0, sizeof(f[0]));
        Now = id[i];
        dfs(1);
        for (int j = k; j <= n; ++j) 
            ans = (ans + f[0][j] * (a[id[i]] - a[id[i - 1]])) % mod;
    }
    cout << ans % mod;
    return 0;
}