【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

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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[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 );
}

 

posted @ 2018-08-13 17:16  Wans_ovo  阅读(115)  评论(0编辑  收藏  举报