【BZOJ】1864: [Zjoi2006]三色二叉树

1864: [Zjoi2006]三色二叉树

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 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

Day1

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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[];
int now, len;

int dp1[][], dp2[][], son[][];
int MA, MI;

int stot, tov[], nex[], h[];

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-] == '' ) {
        now = pos;
        return ;
    }
    if ( s[pos-] == '' ) {
        build ( pos + , pos );
    }
    if ( s[pos-] == '' ) {
        build ( pos + , pos );
        build ( now + , 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][]++] = v;
    }
    int l = son[u][], r = son[u][];
    if ( son[u][] ==  ) {
        dp1[u][] = max ( dp1[l][], dp1[l][] );
        dp1[u][] = max ( dp1[l][], dp1[l][] );
        dp1[u][] = max ( dp1[l][], dp1[l][] ) + ;
        dp2[u][] = min ( dp2[l][], dp2[l][] );
        dp2[u][] = min ( dp2[l][], dp2[l][] );
        dp2[u][] = min ( dp2[l][], dp2[l][] ) + ;
    } else if ( son[u][] ==  ){
        dp1[u][] = max ( dp1[l][] + dp1[r][], dp1[r][] + dp1[l][] );
        dp1[u][] = max ( dp1[l][] + dp1[r][], dp1[r][] + dp1[l][] );
        dp1[u][] = max ( dp1[l][] + dp1[r][], dp1[r][] + dp1[l][] ) + ;
        dp2[u][] = min ( dp2[l][] + dp2[r][], dp2[r][] + dp2[l][] );
        dp2[u][] = min ( dp2[l][] + dp2[r][], dp2[r][] + dp2[l][] );
        dp2[u][] = min ( dp2[l][] + dp2[r][], dp2[r][] + dp2[l][] ) + ;
    } else {
        dp1[u][] = dp1[u][] = ;
        dp1[u][] = ;
        dp2[u][] = dp2[u][] = ;
        dp2[u][] = ;
    }
}

int main ( ) {
    cin >> s;
    len = strlen ( s );
    build ( ,  );
    dfs ( ,  );
    MA = max ( dp1[][], max ( dp1[][], dp1[][] ) );
    MI = min ( dp2[][], min ( dp2[][], dp2[][] ) );
    printf ( "%d %d", MA, MI );
}