BZOJ1369 [Baltic2003]Gem

就是一个简单的树形DP啦~但是问题是到底要几种颜色

我一开始写了25中颜色交上去，发现好慢啊。。。于是做了个死改成了三种交了上去竟然过了。。。过了。。。（要知道2种颜色肯定是不对的啊。。。）

 

 1 /**************************************************************
 2     Problem: 1369
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:12 ms
 7     Memory:1820 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12  
13 using namespace std;
14 const int N = 1e4 + 5;
15 const int inf = 1e9;
16 const int C = 3;
17  
18 struct edge {
19     int next, to;
20     edge() {}
21     edge(int _n, int _t) : next(_n), to(_t) {}
22 } e[N << 1];
23  
24 int n, ans;
25 int first[N], tot;
26 int f[N][21];
27  
28 inline int read() {
29     int x = 0;
30     char ch = getchar();
31     while (ch < '0' || '9' < ch)
32         ch = getchar();
33     while ('0' <= ch && ch <= '9') {
34         x = x * 10 + ch - '0';
35         ch = getchar();
36     }
37     return x;
38 }
39  
40  
41 inline 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 void work(int p, int fa) {
48     int x, i, j, mn;
49     for (i = 1; i <= C; ++i) f[p][i] = i;
50     for (x = first[p]; x; x = e[x].next)
51         if (y != fa) work(y, p);
52     for (i = 1; i <= C; ++i) {
53         for (x = first[p]; x; x = e[x].next)
54             if (y != fa) {
55                 for (j = 1, mn = inf; j <= C; ++j)
56                     if (i != j) mn = min(mn, f[y][j]);
57                 f[p][i] += mn;
58             }
59     }
60 }
61 #undef y
62  
63 int main() {
64     int i;
65     n = read();
66     for (i = 1; i < n; ++i)
67         Add_Edges(read(), read());
68     work(1, 0);
69     for (i = 1, ans = inf; i <= C; ++i)
70         ans = min(ans, f[1][i]);
71     printf("%d\n", ans);
72     return 0;
73 }
View Code

 

posted on 2015-03-06 16:34  Xs酱~  阅读(209)  评论(0)  编辑  收藏  举报