[基环树dp] BZOJ1040

题意

给一个基环森林，每个点有一个权值
求一个点的集合,使集合里面不存在相连的两个点

思路

基环树: 树加一条边使之成环

如果是树就正常树形dp
f[i][0]= Σ{max(f[son(i)][0],f[son(i)][1])} Σ { m a x ( f [ s o n ( i ) ] [ 0 ] , f [ s o n ( i ) ] [ 1 ] ) }
f[i][1]= Σ{f[son(i)][0]} Σ { f [ s o n ( i ) ] [ 0 ] }
现在有一条环边
dfs找到环边
从该边的两个端点出发选择,
1） 强制不选U，V任意，环的贡献为以U做DP的f[U][0]
2） 强制不选V，U任意，环的贡献为以V做DP的f[V][0]

代码

#include 
#include 
#include 
#include 
#include 

using namespace std;
typedef long long ll;

const int INF = 0x7f7f7f7f;
const int maxn = 1e6 + 10;

int n;
ll val[ maxn ];
ll dp[ maxn ][ 2 ];
bool vis[ maxn ];
bool visdp[ maxn ];

struct EDGE {
    int to, nex, id;
} edg[ 2 * maxn ];

int cnt;
ll sol;
int head[ maxn ];
void INit () {
    cnt = 0, sol = 0;
    memset ( dp, 0, sizeof ( dp ) );
    memset ( vis, false, sizeof ( vis ) );
    memset ( visdp, false, sizeof ( visdp ) );
    memset ( head, -1, sizeof ( head ) );
    memset ( edg, 0, sizeof ( edg ) );
}

void AddEdge ( int u, int v ) {
    edg[ cnt ].to = v;
    edg[ cnt ].nex = head[ u ];
    head[ u ] = cnt++;
}

int U, V;
void dfs ( int rt, int fa ) {
    //遍历森林
    vis[ rt ] = true;

    for ( int i = head[ rt ]; i != -1; i = edg[ i ].nex ) {
        int v = edg[ i ].to;
        if ( v == fa )
            continue;
        if ( vis[ v ] ) {
            //在一个子树上找到环边
            U = rt, V = v;
            continue;
        }
        //遍历完整个子树
        dfs ( v, rt );
    }
}

bool fg;
void tree_dp ( int rt ) {
    dp[ rt ][ 1 ] = val[ rt ];
    dp[ rt ][ 0 ] = 0;
    visdp[ rt ] = true;

    for ( int i = head[ rt ]; i != -1; i = edg[ i ].nex ) {
        int v = edg[ i ].to;
        if ( visdp[ v ] )
            continue;
        /*
         * 还是用vis数组比较好用,fa在这样的环里容易循环
         * 不过用fa的话还是不对
            //if ( i == ban || ( i ^ 1 ) == ban  )
            //if ( ( rt == U && v == V ) || ( rt == V && v == U ) )
        if ( fg && v == U ) //从U开始时,V到U的边不能选
            continue;
        if ( !fg && v == V ) //从V开始时,V可以到U,U到V的边不能选
            continue;
        */

        tree_dp ( v );

        dp[ rt ][ 0 ] += max ( dp[ v ][ 0 ], dp[ v ][ 1 ] );
        dp[ rt ][ 1 ] += dp[ v ][ 0 ];
    }
}

int main () {
#ifdef LOCAL
    freopen ( "in", "r", stdin );
    // freopen("out","w",stdout);
#endif
    while ( ~scanf ( "%d", &n ) ) {
        INit ();
        for ( int i = 1, v; i <= n; ++i ) {
            scanf ( "%lld%d", &val[ i ], &v );
            AddEdge ( i, v );
            AddEdge ( v, i );
        }

        for ( int i = 1; i <= n; ++i ) {
            //森林
            if ( !vis[ i ] ) {
                dfs ( i, -1 );

                // fg没用
                //从U出发,U强制不选
                fg = true;
                memset ( visdp, false, sizeof ( visdp ) );
                tree_dp ( U );
                ll tmp = dp[ U ][ 0 ];

                fg = false;
                memset ( visdp, false, sizeof ( visdp ) );
                tree_dp ( V );
                tmp = max ( tmp, dp[ V ][ 0 ] );
                //该子树最大的值
                sol += tmp;
            }
        }
        printf ( "%lld\n", sol );
    }

    return 0;
}