UVALive 2038Strategic game

Posted 2020-08-23

题解:

基础树形dp

dp[u][0] 表示不选u.那么子节点v都要选

dp[u][1]表示选择u，那么子节点v可选可不选

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 = 1600;
int dp[ maxn ][ 2 ];
int n;

int head[ maxn * 2 ], cnt;

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

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

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

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

int main()
{
    while( ~scanf( "%d", &n ) )
    {
        init();
        while( n )
        {
          int u, v, num;
          scanf( "%d:(%d)", &u, &num );
          for( int i = 1; i <= num; i ++  )
          {
              scanf( "%d", &v );
              addedge( u, v );
              addedge( v, u );
          }
          n --;
        }
        dfs( 0, -1 );
        printf( "%d\n", min( dp[ 0 ][ 0 ], dp[ 0 ][ 1 ] ) );
    }
    return 0;
}