POJ 1463 Strategic game( 树形DP )

标签：算法   数学   动态规划   

题意：一颗 N 个节点的树，在某一个节点上放置一个兵则可以守住与它相邻的边。最少放置多少个兵才可以守住所有的边。

#include <cstdio>
#include <deque>

using namespace std;

#define ABANDON 0
#define GET 1

deque< int > graph[2010];
int DP[2010][2];


void DFS( int start, int parent ){

    DP[start][ABANDON] = 0;
    DP[start][GET]     = 1;

    int target;

    if( graph[start].size() == 1 && parent != -1 )
        return;

    for( int i = 0; i < graph[start].size(); ++i ){

        target = graph[start][i];

        if( target == parent )
            continue;

        DFS( target, start );
        DP[start][ABANDON] += DP[target][GET];
        DP[start][GET] += min( DP[target][GET], DP[target][ABANDON] );

    }

}


int main(){

    int nodes;
    int start, roads, target;

    while( scanf( "%d", &nodes ) != EOF ){

        for( int i = 0; i <= nodes; ++i )
            graph[i].clear();

        for( int i = 0; i < nodes; ++i ){

            scanf( "%d:(%d)", &start, &roads );

            while( roads-- ){

                scanf( "%d", &target );
                graph[start].push_back( target );
                graph[target].push_back( start );

            }
        }

        DFS( 0, -1 );

        printf( "%d\n", min( DP[0][ABANDON], DP[0][GET] ) );

    }

    return 0;
}</span>

POJ 1463 Strategic game( 树形DP )

标签：算法   数学   动态规划   

原文地址：http://blog.csdn.net/pandora_madara/article/details/39581285