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POJ 1463 Strategic game( 树形DP )

栏目:互联网时间:2014-09-29 13:26:30

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 ", min( DP[0][ABANDON], DP[0][GET] ) ); } return 0; }</span>


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