HDU4044 GeoDefense

Posted 2020-08-25

题解:

树形dp + 分组背包(两个阶段)

具体解释见代码

代码:

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
#define ll long long

const int maxn = 1010;
const int maxm = 205;
const int INF = 1e9;

int T, n, m;
int dp1[ maxn ][ maxm ], dp2[ maxn ][ maxm ]; 
int attack[ maxn ][ maxm ], num[ maxn ];

//dp1[ u ][ j ]表示u节点不放攻击点，给予费用为j时，其儿子的最小防御值
//dp2[ u ][ j ]表示u节点放或不放攻击点，给予费用为j时，u包括其儿子最小防御值

//动态规划有2步，第一步是求dp1, 然后在dp1的基础上求得dp2
//在dp过程中，因为价格可能为0,所以不能直接用逆向，否则无法保证只能放一个攻击点,需要一个变量t存储
//优化一下攻击方式,attack[ i ][ j ]表示i节点在给予费用为j时的最大攻击力，这样省去很多时间和步骤 


struct Edge
{
    int nxt, v;
}edge[ maxn * 2 ];

int head[ maxn * 2 ], cnt;

void init()
{
    memset( head, -1, sizeof( head ) );
    cnt = 0;
}

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

void dfs( int u ,int fa )
{
    if( head[ u ] == -1 || ( edge[ head[ u ] ].v == fa && edge[ head[ u ] ].nxt == -1 ) )
    {
        for( int i = 0; i <= m; i ++ ) dp2[ u ][ i ] = attack[ u ][ i ];
        return ;
    }
    for( int i = 0; i <= m; i ++ ) dp1[ u ][ i ] = INF;
    for( int i = head[ u ]; i != -1; i = edge[ i ].nxt )
    {
        int v = edge[ i ].v;
        if( v == fa ) continue;
        dfs( v, u );
        for( int j = m; j >= 0; j -- )
        {
            int t = 0;
            for( int k = 0; k <= j; k ++ )
            {
                 t = max( t, min( dp1[ u ][ j - k ], dp2[ v ][ k ] ) );
            }
            dp1[ u ][ j ]= t;
        }
    }
    for( int j = m; j >= 0; j -- )
    {
        int t = 0;
        for( int k = 0; k <= j; k ++ )
        {
            t = max( t, dp1[ u ][ j - k ] + attack[ u ][ k ] );
        }
        dp2[ u ][ j ] = t;
    }
}

int main()
{
    scanf( "%d", &T );
    while( T -- )
    {
        scanf( "%d", &n );
        init();
        memset( attack, 0, sizeof( attack ) );
        for( int i = 1; i < n; i ++ )
        {
            int x, y;
            scanf( "%d%d", &x, &y );
            addEdge( x, y );
            addEdge( y, x );
        }
        scanf( "%d", &m );
        for( int i = 1; i <= n; i ++ )
        {
            scanf( "%d", &num[ i ] );
            int pri, pow;
            for( int j = 0; j < num[ i ]; j ++ )
            {
                scanf( "%d%d", &pri, &pow );
                attack[ i ][ pri ] = max( attack[ i ][ pri ], pow );
            }
        }
        for( int i = 1; i <= n; i ++ )
        for( int j = 1; j <= m; j ++ ) attack[ i ][ j ] = max( attack[ i ][ j ], attack[ i ][ j - 1 ] );
        dfs( 1, -1 );
        printf( "%d\n", dp2[ 1 ][ m ] );
    }
    return 0;
}