BZOJ1907 树的路径覆盖

ydc题解上写着贪心，后来又说是树形dp。。。可惜看不懂(顺便骗三连)

其实就是每个叶子开始拉一条链，从下面一路走上来，遇到能把两条链合起来的就合起来就好了。

/**************************************************************
    Problem: 1907
    User: rausen
    Language: C++
    Result: Accepted
    Time:112 ms
    Memory:1396 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
 
using namespace std;
const int N = 1e4 + 5;
 
struct edge {
    int next, to;
    edge() {}
    edge(int _n, int _t) : next(_n), to(_t) {}
} e[N << 1];
 
struct tree_node {
    int fa, ans, used;
} tr[N];
 
int n;
int first[N], tot;
 
inline int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
void Add_Edges(int x, int y) {
    e[++tot] = edge(first[x], y), first[x] = tot;
    e[++tot] = edge(first[y], x), first[y] = tot;
}
 
#define y e[x].to
int dfs(int p) {
    int x, cnt = 0;
    tr[p].ans = 1, tr[p].used = 0;
    for (x = first[p]; x; x = e[x].next)
        if (y != tr[p].fa) {
            tr[y].fa = p;
            dfs(y);
            tr[p].ans += tr[y].ans;
            if (!tr[y].used) ++cnt;
        }
    if (cnt >= 2) tr[p].ans -= 2, tr[p].used = 1;
    else if (cnt == 1) --tr[p].ans;
}
#undef y
 
int main() {
    int i, T;
    T = read();
    while (T--) {
        n = read(), tot = 0;
        memset(first, 0, sizeof(first));
        for (i = 1; i < n; ++i)
            Add_Edges(read(), read());
        dfs(1);
        printf("%d
", tr[1].ans);
    }
    return 0;
}