bzoj 3522 [Poi2014]Hotel 树形dp

题面

题目传送门

解法

MLE了23333

为什么空间不给256MB，这样我就能开几个5000*5000的数组了

考虑一种树形dp的方法吧

\(f_{i,j}\)表示从点\(i\)向下走\(j\)步并选出两个不在\(i\)同一个子树的方案数，\(g_{i,j}\)表示从\(i\)出发在\(i\)子树外走\(j\)步的方案数

那么答案即为\(\sum f_{i,j}×g_{i,j}\)

注意统计时可能会出现重复的情况

时间复杂度：\(O(n^2)\)

当然，还有一种复杂度为\(O(n^2)\)的方法，直接枚举三个点的中心点，然后暴力dfs统计答案

感觉这个比较容易实现

代码

#include <bits/stdc++.h>
#define N 5010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
    x = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
    int next, num;
} e[N * 3];
int n, cnt, p[N], f[N][N], g[N][N], s[N][N];
long long ans = 0;
void add(int x, int y) {
    e[++cnt] = (Edge) {e[x].next, y};
    e[x].next = cnt;
}
void dpI(int x, int fa) {
    s[x][0] = 1, p[x] = fa; int tot = 0;
    for (int p = e[x].next; p; p = e[p].next) {
        int k = e[p].num;
        if (k == fa) continue;
        dpI(k, x); tot++;
        for (int i = 1; i <= n; i++) {
        	if (tot >= 3) ans += 1ll * f[x][i] * s[k][i - 1];
        	f[x][i] += s[x][i] * s[k][i - 1];
            s[x][i] += s[k][i - 1];
        }
    }
}
void dfs(int x, int fa, int y, int sum) {
	g[y][sum]++;
	for (int p = e[x].next; p; p = e[p].next)
		if (e[p].num != fa) dfs(e[p].num, x, y, sum + 1);
}
int main() {
    read(n); cnt = n;
    for (int i = 1; i < n; i++) {
        int x, y; read(x), read(y);
        add(x, y), add(y, x);
    }
    dpI(1, 0);
    for (int i = 1; i <= n; i++)
    	for (int q = e[i].next; q; q = e[q].next)
    		if (e[q].num == p[i]) dfs(p[i], i, i, 1);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            ans += 1ll * f[i][j] * g[i][j];
    cout << ans << "\n";
    return 0;
}