UVA10859Placing Lampposts

Posted 2020-08-21

题解:

一道很基础的树形DP题目

有趣的是需要在点V1最小的情况下，边两点同时亮的条数足够大

边两点同时亮的条数足够大。等价于边一点同时亮的条数最小V2

构造X = M * V1 + V2

当M足够大时，就把2个变量维护成一个变量....interesting

DP方程

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

+1表示边只有一点亮(excited)

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>
const int INF = 1000000000;
const int maxn = 1050;
const int M = 4000;

int T, n, m;
int cnt;
int head[ maxn * 2 ];
int dp[ maxn ][ 2 ], vis[ maxn ];

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

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

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

void dfs( int u, int fa )
{
   dp[ u ][ 0 ] = 0;
   dp[ u ][ 1 ] = M;
   vis[ u ] = 1;
   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 ][ 0 ] += ( dp[ v ][ 1 ] + 1 );
       dp[ u ][ 1 ] += min( dp[ v ][ 0 ] + 1,dp[ v ][ 1 ] );
   }
}


int main()
{
    scanf( "%d", &T );
    while( T -- )
    {
        scanf( "%d%d", &n, &m );
        init();
        for( int i = 1; i <= m; i ++ )
        {
            int x, y;
            scanf( "%d%d", &x, &y );
            addEdge( x, y );
            addEdge( y, x );
        }
        int ans = 0;
        for( int i = 0; i < n; i ++ )
        {
            if( vis[ i ] ) continue;
            dfs( i, -1 );
            ans += min( dp[ i ][ 0 ], dp[ i ][ 1 ] );
        }
        printf( "%d %d %d\n", ans / M, m - ans % M, ans % M );
    }
    return 0;
}