【树形DP】POJ 2378 Tree Cutting

通道：http://poj.org/problem?id=2378

题意：

     给一颗n个结点的树，节点编号为1~n，把删除一个节点之后，
     剩下的分支中节点数量最多的数量不大于总数量一半的编号全部按顺序输出

，和上题类似

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

typedef pair<int, int >PII;
typedef long long int64;
const int INF = 0x3f3f3f3f;

const int MAXN = 10010;
int tot[MAXN];
int f[MAXN], minx;

namespace Adj {
    int size, head[MAXN];
    struct Node{
        int v, next; 
    }E[MAXN*2];
    inline void initAdj() {
        size = 0;
        memset(head, -1, sizeof(head));
    }
    inline void addEdge(int u, int v) {
        E[size].v = v;
        E[size].next = head[u];
        head[u] = size++;
    }
}
using namespace Adj;

int n;

int dfs(int u, int fa) {

    tot[u] = 1;
    // count
    for (int e = head[u]; e != -1; e = E[e].next) {
        int v = E[e].v;
        if (v == fa) continue;
        tot[u] += dfs(v, u);
    }
    // 计算答案
    int& ans = f[u] = n - tot[u];
    for (int e = head[u]; e != -1; e = E[e].next) {
        int v = E[e].v;
        if (v == fa) continue;
        ans = max(ans, tot[v]);
    }
    minx = min(minx, ans);
    return tot[u];
}

int main(){

    while (~scanf("%d", &n)) {

        initAdj();
        for (int i = 0; i < n - 1; ++i) {
            int u, v;
            scanf("%d%d", &u, &v);
            addEdge(u, v);
            addEdge(v, u);
        }

        dfs(1, -1);

        for (int i = 1; i <= n; ++i) 
            if (f[i] <= n/2) {
                printf("%d\n", i);
            }
        puts("");
    }
    return 0;
}