【BZOJ】1864: [Zjoi2006]三色二叉树
1864: [Zjoi2006]三色二叉树
Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 1295 Solved: 961
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Description
Input
仅有一行,不超过500000个字符,表示一个二叉树序列。
Output
输出文件也只有一行,包含两个数,依次表示最多和最少有多少个点能够被染成绿色。
Sample Input
1122002010
Sample Output
5 2
HINT
Source
其实是一道基础的树规辣~一开始觉得难点可能在建边上,可是建边也很好搞是怎么肥四??(其实就是一道水题
建了边直接dfs往上dp就可以辣!
#include<iostream> #include<cstdio> #include<cstring> using namespace std; char s[500005]; int now, len; int dp1[1000005][3], dp2[1000005][3], son[1000005][3]; int MA, MI; int stot, tov[2000005], nex[2000005], h[1000005]; void add ( int u, int v ) { tov[++stot] = v; nex[stot] = h[u]; h[u] = stot; } void build ( int pos, int f ) { add ( f, pos ); add ( pos, f ); if ( s[pos-1] == '0' ) { now = pos; return ; } if ( s[pos-1] == '1' ) { build ( pos + 1, pos ); } if ( s[pos-1] == '2' ) { build ( pos + 1, pos ); build ( now + 1, pos ); } } void dfs ( int u, int f ) { for ( int i = h[u]; i; i = nex[i] ) { int v = tov[i]; if ( v == f ) continue; dfs ( v, u ); son[u][son[u][2]++] = v; } int l = son[u][0], r = son[u][1]; if ( son[u][2] == 1 ) { dp1[u][0] = max ( dp1[l][1], dp1[l][2] ); dp1[u][1] = max ( dp1[l][0], dp1[l][2] ); dp1[u][2] = max ( dp1[l][1], dp1[l][0] ) + 1; dp2[u][0] = min ( dp2[l][1], dp2[l][2] ); dp2[u][1] = min ( dp2[l][0], dp2[l][2] ); dp2[u][2] = min ( dp2[l][1], dp2[l][0] ) + 1; } else if ( son[u][2] == 2 ){ dp1[u][0] = max ( dp1[l][1] + dp1[r][2], dp1[r][1] + dp1[l][2] ); dp1[u][1] = max ( dp1[l][0] + dp1[r][2], dp1[r][0] + dp1[l][2] ); dp1[u][2] = max ( dp1[l][1] + dp1[r][0], dp1[r][1] + dp1[l][0] ) + 1; dp2[u][0] = min ( dp2[l][1] + dp2[r][2], dp2[r][1] + dp2[l][2] ); dp2[u][1] = min ( dp2[l][0] + dp2[r][2], dp2[r][0] + dp2[l][2] ); dp2[u][2] = min ( dp2[l][1] + dp2[r][0], dp2[r][1] + dp2[l][0] ) + 1; } else { dp1[u][0] = dp1[u][1] = 0; dp1[u][2] = 1; dp2[u][0] = dp2[u][1] = 0; dp2[u][2] = 1; } } int main ( ) { cin >> s; len = strlen ( s ); build ( 1, 0 ); dfs ( 1, 0 ); MA = max ( dp1[1][0], max ( dp1[1][1], dp1[1][2] ) ); MI = min ( dp2[1][0], min ( dp2[1][1], dp2[1][2] ) ); printf ( "%d %d", MA, MI ); }