UVa10817

题目大意

某校有n个教师和m个求职者，已知每人的工资和能教的课程集合，要求支付最少的工资使得每门课都至少有两名教师教学。在职教师必须招聘。

思路

这题不太好想,搞了很久。。

f[s1][s2]： s1表示课程集合{ s1 }都至少有一个教师教的情况。

               s2表示课程集合{ s2 }都至少有两个教师教的情况。

每个求职者的pi, 对于每个求职者，要么选，要么不选，就是01背包问题。

对于s1，s2，可以根据当前枚举到的求职者课程即可，可推出下一个状态：

nextS1 = p[i] | s1,

nextS2 = (p[i] & s1) | s2

f[nextS1][nextS2] = min(f[nextS1][nextS2], f[s1][s2] + p[i])

代码

<a target=_blank id="L1" href="http://blog.youkuaiyun.com/shuangde800/article/details/9776645#L1" rel="#L1" style="color: rgb(102, 102, 102); text-decoration: none;"> 1</a> <a target=_blank id="L2" href="http://blog.youkuaiyun.com/shuangde800/article/details/9776645#L2" rel="#L2" style="color: rgb(102, 102, 102); text-decoration: none;"> 2</a> <a target=_blank id="L3" href="http://blog.youkuaiyun.com/shuangde800/article/details/9776645#L3" rel="#L3" style="color: rgb(102, 102, 102); text-decoration: none;"> 3</a> <a target=_blank id="L4" href="http://blog.youkuaiyun.com/shuangde800/article/details/9776645#L4" rel="#L4" style="color: rgb(102, 102, 102); text-decoration: none;"> 4</a> <a target=_blank id="L5" href="http://blog.youkuaiyun.com/shuangde800/article/details/9776645#L5" rel="#L5" style="color: rgb(102, 102, 102); text-decoration: none;"> 5</a> <a target=_blank id="L6" href="http://blog.youkuaiyun.com/shuangde800/article/details/9776645#L6" rel="#L6" style="color: rgb(102, 102, 102); 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/**========================================== * This is a solution for ACM/ICPC problem * * @problem: UVA 10817 - Headmaster's Headache * @type: dp * @author: shuangde * @blog: blog.youkuaiyun.com/shuangde800 * @email: zengshuangde@gmail.com *===========================================*/ #include<iostream> #include<cstdio> #include<algorithm> #include<vector> #include<queue> #include<cmath> #include<cstring> using namespace std ; typedef long long int64 ; const int INF = 0x3f3f3f3f ; const double PI = acos ( - 1.0 ); const int MAXN = 150 ; int courseNum , m , n , sum ; int maxState ; int f [ 1 << 8 ][ 1 << 8 ]; int p [ MAXN ], c [ MAXN ], cnt [ 10 ]; int dp ( int st1 , int st2 ){ memset ( f , 0x3f , sizeof ( f )); f [ st1 ][ st2 ] = sum ; for ( int i = m + 1 ; i <= n + m ; ++ i ){ for ( int s1 = maxState ; s1 >= 0 ; -- s1 ){ for ( int s2 = maxState ; s2 >= 0 ; -- s2 ){ if ( f [ s1 ][ s2 ] >= INF ) continue ; int st1 = ( p [ i ] | s1 ); int st2 = ( p [ i ] & s1 ) | s2 ; f [ st1 ][ st2 ] = min ( f [ st1 ][ st2 ], f [ s1 ][ s2 ] + c [ i ]); } } } return f [ maxState ][ maxState ]; } int main (){ char str [ 1000 ]; while ( ~ scanf ( "%d%d%d" , & courseNum , & m , & n ) && courseNum ){ maxState = ( 1 << courseNum ) - 1 ; sum = 0 ; int st1 = 0 , st2 = 0 ; memset ( cnt , 0 , sizeof ( cnt )); for ( int i = 1 ; i <= m + n ; ++ i ){ scanf ( "%d" , & c [ i ]); gets ( str ); p [ i ] = 0 ; for ( int j = 0 ; j < strlen ( str ); ++ j ){ if ( isdigit ( str [ j ])){ int num = str [ j ] - '0' ; p [ i ] |= 1 << ( num - 1 ); if ( i <= m ) ++ cnt [ num - 1 ]; } } if ( i <= m ){ sum += c [ i ]; st1 |= p [ i ]; } } for ( int i = 0 ; i < courseNum ; ++ i ) if ( cnt [ i ] > 1 ) st2 |= ( 1 << i ); printf ( "%d \n " , dp ( st1 , st2 )); } return 0 ; }