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。

 

 1 /**************************************************************
 2     Problem: 1907
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:112 ms
 7     Memory:1396 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12  
13 using namespace std;
14 const int N = 1e4 + 5;
15  
16 struct edge {
17     int next, to;
18     edge() {}
19     edge(int _n, int _t) : next(_n), to(_t) {}
20 } e[N << 1];
21  
22 struct tree_node {
23     int fa, ans, used;
24 } tr[N];
25  
26 int n;
27 int first[N], tot;
28  
29 inline int read() {
30     int x = 0;
31     char ch = getchar();
32     while (ch < '0' || '9' < ch)
33         ch = getchar();
34     while ('0' <= ch && ch <= '9') {
35         x = x * 10 + ch - '0';
36         ch = getchar();
37     }
38     return x;
39 }
40  
41 void Add_Edges(int x, int y) {
42     e[++tot] = edge(first[x], y), first[x] = tot;
43     e[++tot] = edge(first[y], x), first[y] = tot;
44 }
45  
46 #define y e[x].to
47 int dfs(int p) {
48     int x, cnt = 0;
49     tr[p].ans = 1, tr[p].used = 0;
50     for (x = first[p]; x; x = e[x].next)
51         if (y != tr[p].fa) {
52             tr[y].fa = p;
53             dfs(y);
54             tr[p].ans += tr[y].ans;
55             if (!tr[y].used) ++cnt;
56         }
57     if (cnt >= 2) tr[p].ans -= 2, tr[p].used = 1;
58     else if (cnt == 1) --tr[p].ans;
59 }
60 #undef y
61  
62 int main() {
63     int i, T;
64     T = read();
65     while (T--) {
66         n = read(), tot = 0;
67         memset(first, 0, sizeof(first));
68         for (i = 1; i < n; ++i)
69             Add_Edges(read(), read());
70         dfs(1);
71         printf("%d\n", tr[1].ans);
72     }
73     return 0;
74 }
View Code

 

posted on 2015-02-23 19:27  Xs酱~  阅读(267)  评论(0)  编辑  收藏  举报