UVA10817 Headmaster's Headache 状态压缩的01背包
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题目地址:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1758
题目意思:
某校有n个教师和m个求职者。已知每人的工资和能教授的课程集合,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:m个求职者只有招和不招两种情况,0,1背包,只不过背包容量变抽象了而已,“每门课都至少有两名教师教学”是容量,如何表示从0容量转移到最终总容量需要细化状态,再定义一个"当前只有一个人教的课程集合"为另一维的状态就可以实现状态转移了
//765 ms C++ 4.8.2 1378 #include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int dp[125][1<<8][1<<8];int v[125];int st[125];int s,n,m;int main(){ while(scanf("%d%d%d",&s,&n,&m)&&s) { for(int i=0;i<n+m;i++) { scanf("%d",&v[i]); int t=0; while(1) { int tmp; if(getchar()=='\n') break; scanf("%d",&tmp); t|=(1<<(tmp-1)); } st[i]=t; } memset(dp,0x3f,sizeof(dp)); dp[0][0][0]=0; for(int i=0;i<m+n;i++) for(int s1=0;s1<(1<<s);s1++) for(int s2=0;s2<(1<<s);s2++) { if(s1&s2) continue; if( i>=n ) dp[i+1][s1][s2] =min(dp[i+1][s1][s2] , dp[i][s1][s2]); int t=( (1<<s)-1 )^(s1|s2); int S2=s2|(s1&st[i]); int S1=(s1|(st[i]&t))^(s1&st[i]); dp[i+1][S1][S2]=min(dp[i+1][S1][S2], dp[i][s1][s2]+v[i] ); } int ans=inf; for(int s1=0; s1<(1<<s) ;s1++) { if(s1&((1<<s)-1)) continue; ans=min(ans,dp[m+n][s1][(1<<s)-1]); } printf("%d\n",ans); } return 0;}
由于枚举状态会产生很多不符合条件的状态,所以用记忆化搜索可以大大加速:
//AcceptedC++0.569#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int dp[125][(1<<8)][(1<<8)];int n,m,s;int v[125];int st[125];int mrize(int i,int s0,int s1,int s2){ if(i==m+n) return s2==(1<<s)-1 ? 0:inf; int &ans =dp[i][s1][s2]; if( ans>=0 ) return ans; ans=inf; if(i>=n) ans=min(ans,mrize(i+1,s0,s1,s2)); s2|=(s1&st[i]); s1 =(s1|(s0&st[i]))^(s1&st[i]); s0^=(s0&st[i]); ans=min(ans,mrize(i+1,s0,s1,s2)+v[i]); return ans;}int main(){ while(scanf("%d%d%d",&s,&n,&m)&&s) { memset(dp,-1,sizeof(dp)); for(int i=0;i<n+m;i++) { scanf("%d",&v[i]); int t=0; while(1) { int tmp; if(getchar()=='\n') break; scanf("%d",&tmp); t|=(1<<(tmp-1)); } st[i]=t; } printf("%d\n",mrize(0,(1<<s)-1,0,0)); } return 0;}
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