poj 1463 树形dp入门

很简单的树形dp题目，转移方程是：

     dp[u][0] += dp[v][1];
     dp[u][1] += min( dp[v][0], dp[v][1] );

其中u是v的父亲节点。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 1500;
bool flag[N];
int head[N];
int dp[N][2];
int e, n;

void init()
{
    e = 0;
    memset( flag, true, sizeof(flag) );
    memset( head, -1, sizeof(head) );
}

struct Edge
{
    int v, next;
} edge[N];

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

void dfs( int u )
{
    dp[u][0] = 0;
    dp[u][1] = 1;
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        dfs(v);
        dp[u][0] += dp[v][1];
        dp[u][1] += min( dp[v][0], dp[v][1] );
    }
}

int main ()
{
    while ( scanf("%d", &n) != EOF )
    {
        init();
        for ( int i = 0; i < n; i++ )
        {
            int u, v, m;
            scanf("%d:(%d)", &u, &m);
            while ( m-- )
            {
                scanf("%d", &v);
                addEdge( u, v );
                flag[v] = false;
            }
        }
        int root;
        for ( int i = 0; i < n; i++ )
        {
            if ( flag[i] )
            {
                root = i;
                break;
            }
        }
        dfs(root);
        int ans = min( dp[root][0], dp[root][1] );
        printf("%d
", ans);
    }
    return 0;
}

也可以用二分图来做。