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 ", 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 }