hdu 4799 LIKE vs CANDLE

题解：这道题是比较明显的树状dp,状态方程也挺好想的，主要是要处理反转和不反转的关系，可以在全局定义一个标志位，来记录这棵子树下面的反转情况，然后定义子树的遍历情况。dp[i][j]是说i这个节点是否转变，转的话j为0,不转为1.

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <map>const int maxn = 50000 + 100;using namespace std;struct node{    int v, next;}p[maxn * 2];int vis[maxn];int head[maxn];int s[maxn];int v[maxn];int e;int flag;int dp[maxn][3];int n, a, b;int x, y;void init(){    memset(head , -1, sizeof(head));    memset(vis, 0, sizeof(vis));    memset(dp, 0, sizeof(dp));    e = 0;    flag = 0;    }void dfs(int x){    vis[x] = 1;    if(s[x]) flag = flag ^ 1;    if(flag) v[x] = -v[x];    dp[x][0] = v[x];    dp[x][1] = -v[x];    int sum;    for(int i = head[x]; i != -1; i = p[i].next)    {        int t = p[i].v;        if(vis[t]) continue;        dfs(t);        if(s[t])        sum = b;        else        sum = a;        dp[x][1] += max(dp[t][1], dp[t][0] - sum);        dp[x][0] += max(dp[t][0], dp[t][1] - sum);    }    if(s[x]) flag = flag ^ 1;    }int main(){    while(~scanf("%d%d%d", &n, &a, &b))    {        init();        for(int i = 1; i <= n; i ++)        {            scanf("%d %d %d %d", &v[i], &x, &s[i], &y);            if(y == 1)             v[i] = -v[i];             p[e].v = i;             p[e].next = head[x];             head[x] = e ++;        }        dfs(0);        if(dp[0][0] < 0)        {            printf("HAHAHAOMG\n");        }        else        printf("%d\n", dp[0][0]);    }}