HDU 1520 anniversary party tree DP

Link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1520

Question: Give a workplace relationship diagram. Each person has his own value and each person has only one leader. The person I want to invite cannot be a person or his immediate leader, ask the maximum number of people who can be invited.

Idea: the entry question is that it is interesting to apply simple DP to a data structure like a tree. It is suggested that forest DP is more appropriate than forest DP. The first step is to build a tree. The root node is a node without Father's Day. For a node u on the tree, DP [u] [0] represents the maximum value that can be obtained without selecting this node, DP [u] [1] indicates the maximum value that can be obtained by selecting this node.

State transition equation: DP [u] [0] = sum (max (DP [v] [0], DP [v] [1]). If the current node is not selected, each of its subnodes can be selected or not selected.

DP [u] [1] = sum (DP [v] [0]). If you select the current node, each of its subnodes cannot be selected.

Code:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<ctype.h>#include<algorithm>#include<string>#define PI acos(-1.0)#define maxn 6005#define INF 0x7ffffffftypedef long long LL;using namespace std;int score[maxn],head[maxn],top,dp[maxn][2];bool from[maxn];void init(){    memset(head,-1,sizeof(head));    memset(from,0,sizeof(from));    memset(dp,0,sizeof(dp));    top=0;}struct Edge{    int v;    int next;} edge[maxn];int add_edge(int u,int v){    edge[top].v=v;    edge[top].next=head[u];    head[u]=top++;}void dfs(int u){    for(int i=head[u]; i!=-1; i=edge[i].next)    {        int v=edge[i].v;        dfs(v);        dp[u][1]+=dp[v][0];        dp[u][0]+=max(dp[v][1],dp[v][0]);    }    dp[u][1]+=score[u];}int main(){    int tot,u,v;    while(~scanf("%d",&tot))    {        init();        for(int i=1; i<=tot; i++)            scanf("%d",&score[i]);        while(scanf("%d%d",&u,&v))        {            if(u==0&&v==0)                break;            add_edge(v,u);            from[u]=true;        }        for(int i=1; i<=tot; i++)        {            if(!from[i])                dfs(i);        }        int ans=0;        for(int i=1; i<=tot; i++)        {            if(from[i]==0)                ans+=max(dp[i][0],dp[i][1]);        }        printf("%d\n",ans);    }    return 0;}

HDU 1520 anniversary party tree DP