简单地树形DP

我们用f,g表示最大、最小值,0,1,2表示颜色然后直接推

递推公式请见程序233

 

 1 /**************************************************************
 2     Problem: 1864
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:24 ms
 7     Memory:19244 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12   
13 #define Rep(i, n) for (i = 0; i < n; ++i)
14 using namespace std;
15 const int inf = (int) 1e9;
16 const int N = 500005;
17   
18 int cnt, s[N][3], f[N][3], g[N][3];
19   
20 void dfs(int p) {
21     int i, j, k;
22     s[p][0] = getchar() - '0';
23     if (!s[p][0]) {
24         f[p][0] = g[p][0] = 1;
25         return;
26     }
27     for (i = 1; i <= s[p][0]; ++i) {
28         s[p][i] = ++cnt;
29         dfs(cnt);
30     }
31     g[p][0] = g[p][1] = g[p][2] = inf;
32     if (s[p][0] == 1) {
33         Rep(i, 3) Rep(j, 3)
34             if (i != j)
35                 f[p][i] = max(f[p][i], f[s[p][1]][j] + !i),
36                 g[p][i] = min(g[p][i], g[s[p][1]][j] + !i);
37     } else {
38         Rep(i, 3) Rep(j, 3) Rep(k, 3)
39             if (i != j && j != k && i != k)
40                 f[p][i] = max(f[p][i], f[s[p][1]][j] + f[s[p][2]][k] + !i),
41                 g[p][i] = min(g[p][i], g[s[p][1]][j] + g[s[p][2]][k] + !i);
42     }
43 }
44   
45 int main() {
46     dfs(cnt = 1);
47     printf("%d %d\n", max(max(f[1][0], f[1][1]), f[1][2]), min(min(g[1][0], g[1][1]), g[1][2]));
48     return 0;
49 }
View Code

 

By Xs酱~ 转载请说明 博客地址:http://www.cnblogs.com/rausen
posted on 2014-12-03 21:52  Xs酱~  阅读(223)  评论(0编辑  收藏  举报